Max Min Word Problems

Our approach to max min word problems is modeled after ourapproach to related rates word problems. We will

1. draw a sketch of the situation;

2. label every quantity that can vary with a letter;

3. write down the information of the problem in terms of thoseletters;

4. write down other relevant facts;

5. restate and solve the problem;

6. guarantee that our solution is as claimed.

Max Min Word Problems

Our approach to max min word problems is modeled after ourapproach to related rates word problems. We will

1. draw a sketch of the situation;

2. label every quantity that can vary with a letter;

3. write down the information of the problem in terms of thoseletters;

4. write down other relevant facts;

5. restate and solve the problem;

6. guarantee that our solution is as claimed.

Max Min Word Problems

Our approach to max min word problems is modeled after ourapproach to related rates word problems. We will

1. draw a sketch of the situation;

2. label every quantity that can vary with a letter;

3. write down the information of the problem in terms of thoseletters;

4. write down other relevant facts;

5. restate and solve the problem;

6. guarantee that our solution is as claimed.

Max Min Word Problems

Our approach to max min word problems is modeled after ourapproach to related rates word problems. We will

1. draw a sketch of the situation;

2. label every quantity that can vary with a letter;

3. write down the information of the problem in terms of thoseletters;

4. write down other relevant facts;

5. restate and solve the problem;

6. guarantee that our solution is as claimed.

Max Min Word Problems

Our approach to max min word problems is modeled after ourapproach to related rates word problems. We will

1. draw a sketch of the situation;

2. label every quantity that can vary with a letter;

3. write down the information of the problem in terms of thoseletters;

4. write down other relevant facts;

5. restate and solve the problem;

6. guarantee that our solution is as claimed.

Max Min Word Problems

Our approach to max min word problems is modeled after ourapproach to related rates word problems. We will

1. draw a sketch of the situation;

2. label every quantity that can vary with a letter;

3. write down the information of the problem in terms of thoseletters;

4. write down other relevant facts;

5. restate and solve the problem;

6. guarantee that our solution is as claimed.

Max Min Word Problems

Our approach to max min word problems is modeled after ourapproach to related rates word problems. We will

1. draw a sketch of the situation;

2. label every quantity that can vary with a letter;

3. write down the information of the problem in terms of thoseletters;

4. write down other relevant facts;

5. restate and solve the problem;

6. guarantee that our solution is as claimed.

Biggest Garden

Example 1. Problem. A farmer plans to make a rectangulargarden. One side will be against a long barn. He has 100 ft offencing that he will use to surround the other three sides. Whatare the dimensions of the garden of maximum area?

y

x

x + 2y = 100Area = A = xy =(100− 2y)y =100y − 2y2.

We wish tofind the maximum valueof A and the values of xand y that produce thatmaximum value of A.

x ≥ 0 and y ≥ 0. Since x = 100− 2y , we see that 100− 2y ≥ 0.So 100 ≥ 2y and 50 ≥ y . That is, 0 ≤ y ≤ 50.

Biggest Garden

Example 1. Problem. A farmer plans to make a rectangulargarden. One side will be against a long barn. He has 100 ft offencing that he will use to surround the other three sides. Whatare the dimensions of the garden of maximum area?

y

x

x + 2y = 100Area = A = xy =(100− 2y)y =100y − 2y2.

We wish tofind the maximum valueof A and the values of xand y that produce thatmaximum value of A.

x ≥ 0 and y ≥ 0. Since x = 100− 2y , we see that 100− 2y ≥ 0.So 100 ≥ 2y and 50 ≥ y . That is, 0 ≤ y ≤ 50.

Biggest Garden

Example 1. Problem. A farmer plans to make a rectangulargarden. One side will be against a long barn. He has 100 ft offencing that he will use to surround the other three sides. Whatare the dimensions of the garden of maximum area?

y

x

x + 2y = 100

Area = A = xy =(100− 2y)y =100y − 2y2.

We wish tofind the maximum valueof A and the values of xand y that produce thatmaximum value of A.

x ≥ 0 and y ≥ 0. Since x = 100− 2y , we see that 100− 2y ≥ 0.So 100 ≥ 2y and 50 ≥ y . That is, 0 ≤ y ≤ 50.

Biggest Garden

Example 1. Problem. A farmer plans to make a rectangulargarden. One side will be against a long barn. He has 100 ft offencing that he will use to surround the other three sides. Whatare the dimensions of the garden of maximum area?

y

x

x + 2y = 100Area = A = xy =(100− 2y)y =100y − 2y2.

We wish tofind the maximum valueof A and the values of xand y that produce thatmaximum value of A.

x ≥ 0 and y ≥ 0. Since x = 100− 2y , we see that 100− 2y ≥ 0.So 100 ≥ 2y and 50 ≥ y . That is, 0 ≤ y ≤ 50.

Biggest Garden

Example 1. Problem. A farmer plans to make a rectangulargarden. One side will be against a long barn. He has 100 ft offencing that he will use to surround the other three sides. Whatare the dimensions of the garden of maximum area?

y

x

x + 2y = 100Area = A = xy =(100− 2y)y =100y − 2y2.

We wish tofind the maximum valueof A and the values of xand y that produce thatmaximum value of A.

x ≥ 0 and y ≥ 0. Since x = 100− 2y , we see that 100− 2y ≥ 0.So 100 ≥ 2y and 50 ≥ y . That is, 0 ≤ y ≤ 50.

Biggest Garden

Example 1. Problem. A farmer plans to make a rectangulargarden. One side will be against a long barn. He has 100 ft offencing that he will use to surround the other three sides. Whatare the dimensions of the garden of maximum area?

y

x

x + 2y = 100Area = A = xy =(100− 2y)y =100y − 2y2.

We wish tofind the maximum valueof A and the values of xand y that produce thatmaximum value of A.

x ≥ 0 and y ≥ 0.

Since x = 100− 2y , we see that 100− 2y ≥ 0.So 100 ≥ 2y and 50 ≥ y . That is, 0 ≤ y ≤ 50.

Biggest Garden

Example 1. Problem. A farmer plans to make a rectangulargarden. One side will be against a long barn. He has 100 ft offencing that he will use to surround the other three sides. Whatare the dimensions of the garden of maximum area?

y

x

x + 2y = 100Area = A = xy =(100− 2y)y =100y − 2y2.

We wish tofind the maximum valueof A and the values of xand y that produce thatmaximum value of A.

x ≥ 0 and y ≥ 0. Since x = 100− 2y , we see that 100− 2y ≥ 0.

So 100 ≥ 2y and 50 ≥ y . That is, 0 ≤ y ≤ 50.

Biggest Garden

Example 1. Problem. A farmer plans to make a rectangulargarden. One side will be against a long barn. He has 100 ft offencing that he will use to surround the other three sides. Whatare the dimensions of the garden of maximum area?

y

x

x + 2y = 100Area = A = xy =(100− 2y)y =100y − 2y2.

We wish tofind the maximum valueof A and the values of xand y that produce thatmaximum value of A.

x ≥ 0 and y ≥ 0. Since x = 100− 2y , we see that 100− 2y ≥ 0.So 100 ≥ 2y and 50 ≥ y .

That is, 0 ≤ y ≤ 50.

Biggest Garden

Example 1. Problem. A farmer plans to make a rectangulargarden. One side will be against a long barn. He has 100 ft offencing that he will use to surround the other three sides. Whatare the dimensions of the garden of maximum area?

y

x

x + 2y = 100Area = A = xy =(100− 2y)y =100y − 2y2.

We wish tofind the maximum valueof A and the values of xand y that produce thatmaximum value of A.

x ≥ 0 and y ≥ 0. Since x = 100− 2y , we see that 100− 2y ≥ 0.So 100 ≥ 2y and 50 ≥ y . That is, 0 ≤ y ≤ 50.

Biggest Garden

So we use our techniques for finding the maximum value of afunction on a closed interval when the function is continuous there.

These apply since 100y − 2y2 is a polynomial and so continuouseverywhere.dA

dy= 100− 4y .

A is differentiable everywhere, anddA

dy= 0 iff y = 25.

If y = 0, then A = 0.If y = 25, then A = 100 · 25− 2 · 252 = (100− 50)25 > 0.If y = 50, then x = 100− 2 · 50 = 0, and A = 0 · 50 = 0.So the maximum area occurs with y = 25 and x = 100− 2y = 50.

Biggest Garden

So we use our techniques for finding the maximum value of afunction on a closed interval when the function is continuous there.These apply since 100y − 2y2 is a polynomial and so continuouseverywhere.

dA

dy= 100− 4y .

A is differentiable everywhere, anddA

dy= 0 iff y = 25.

If y = 0, then A = 0.If y = 25, then A = 100 · 25− 2 · 252 = (100− 50)25 > 0.If y = 50, then x = 100− 2 · 50 = 0, and A = 0 · 50 = 0.So the maximum area occurs with y = 25 and x = 100− 2y = 50.

Biggest Garden

So we use our techniques for finding the maximum value of afunction on a closed interval when the function is continuous there.These apply since 100y − 2y2 is a polynomial and so continuouseverywhere.dA

dy= 100− 4y .

A is differentiable everywhere, anddA

dy= 0 iff y = 25.

If y = 0, then A = 0.If y = 25, then A = 100 · 25− 2 · 252 = (100− 50)25 > 0.If y = 50, then x = 100− 2 · 50 = 0, and A = 0 · 50 = 0.So the maximum area occurs with y = 25 and x = 100− 2y = 50.

Biggest Garden

So we use our techniques for finding the maximum value of afunction on a closed interval when the function is continuous there.These apply since 100y − 2y2 is a polynomial and so continuouseverywhere.dA

dy= 100− 4y .

A is differentiable everywhere, anddA

dy= 0 iff y = 25.

If y = 0, then A = 0.If y = 25, then A = 100 · 25− 2 · 252 = (100− 50)25 > 0.If y = 50, then x = 100− 2 · 50 = 0, and A = 0 · 50 = 0.So the maximum area occurs with y = 25 and x = 100− 2y = 50.

Biggest Garden

So we use our techniques for finding the maximum value of afunction on a closed interval when the function is continuous there.These apply since 100y − 2y2 is a polynomial and so continuouseverywhere.dA

dy= 100− 4y .

A is differentiable everywhere, anddA

dy= 0 iff y = 25.

If y = 0, then A = 0.

If y = 25, then A = 100 · 25− 2 · 252 = (100− 50)25 > 0.If y = 50, then x = 100− 2 · 50 = 0, and A = 0 · 50 = 0.So the maximum area occurs with y = 25 and x = 100− 2y = 50.

Biggest Garden

So we use our techniques for finding the maximum value of afunction on a closed interval when the function is continuous there.These apply since 100y − 2y2 is a polynomial and so continuouseverywhere.dA

dy= 100− 4y .

A is differentiable everywhere, anddA

dy= 0 iff y = 25.

If y = 0, then A = 0.If y = 25, then A = 100 · 25− 2 · 252 = (100− 50)25 > 0.

If y = 50, then x = 100− 2 · 50 = 0, and A = 0 · 50 = 0.So the maximum area occurs with y = 25 and x = 100− 2y = 50.

Biggest Garden

So we use our techniques for finding the maximum value of afunction on a closed interval when the function is continuous there.These apply since 100y − 2y2 is a polynomial and so continuouseverywhere.dA

dy= 100− 4y .

A is differentiable everywhere, anddA

dy= 0 iff y = 25.

If y = 0, then A = 0.If y = 25, then A = 100 · 25− 2 · 252 = (100− 50)25 > 0.If y = 50, then x = 100− 2 · 50 = 0, and A = 0 · 50 = 0.

So the maximum area occurs with y = 25 and x = 100− 2y = 50.

Biggest Garden

So we use our techniques for finding the maximum value of afunction on a closed interval when the function is continuous there.These apply since 100y − 2y2 is a polynomial and so continuouseverywhere.dA

dy= 100− 4y .

A is differentiable everywhere, anddA

dy= 0 iff y = 25.

If y = 0, then A = 0.If y = 25, then A = 100 · 25− 2 · 252 = (100− 50)25 > 0.If y = 50, then x = 100− 2 · 50 = 0, and A = 0 · 50 = 0.So the maximum area occurs with y = 25 and x = 100− 2y = 50.

Closest Point

Example 2. Problem. Find the point on y = x2 closest to thepoint 〈3, 0〉.

d

xK3 K2 K1 0 1 2 3

y

K1

1

2

3

4

5

6

We let 〈x , y〉 be a point on the curve, and we letd be the distance between 〈x , y〉 and 〈3, 0〉. Sod =

√(x − 3)2 + (y − 0)2 =

√(x − 3)2 + y2.

We want to find the point 〈x , y〉that gives the minimum value for the distance d .

RemarkWe let s = d2. So s = (x − 3)2 + y2. Theminimum value for s and the minimum value ford will occur at the same point. (Why?) So wewill work with s instead of d . This approach isfrequently useful in max min distance problems.

Closest Point

Example 2. Problem. Find the point on y = x2 closest to thepoint 〈3, 0〉.

d

xK3 K2 K1 0 1 2 3

y

K1

1

2

3

4

5

6

We let 〈x , y〉 be a point on the curve, and we letd be the distance between 〈x , y〉 and 〈3, 0〉.

Sod =

√(x − 3)2 + (y − 0)2 =

√(x − 3)2 + y2.

We want to find the point 〈x , y〉that gives the minimum value for the distance d .

RemarkWe let s = d2. So s = (x − 3)2 + y2. Theminimum value for s and the minimum value ford will occur at the same point. (Why?) So wewill work with s instead of d . This approach isfrequently useful in max min distance problems.

Closest Point

Example 2. Problem. Find the point on y = x2 closest to thepoint 〈3, 0〉.

d

xK3 K2 K1 0 1 2 3

y

K1

1

2

3

4

5

6

We let 〈x , y〉 be a point on the curve, and we letd be the distance between 〈x , y〉 and 〈3, 0〉. Sod =

√(x − 3)2 + (y − 0)2 =

√(x − 3)2 + y2.

We want to find the point 〈x , y〉that gives the minimum value for the distance d .

RemarkWe let s = d2. So s = (x − 3)2 + y2. Theminimum value for s and the minimum value ford will occur at the same point. (Why?) So wewill work with s instead of d . This approach isfrequently useful in max min distance problems.

Closest Point

Example 2. Problem. Find the point on y = x2 closest to thepoint 〈3, 0〉.

d

xK3 K2 K1 0 1 2 3

y

K1

1

2

3

4

5

6

We let 〈x , y〉 be a point on the curve, and we letd be the distance between 〈x , y〉 and 〈3, 0〉. Sod =

√(x − 3)2 + (y − 0)2 =

√(x − 3)2 + y2.

We want to find the point 〈x , y〉that gives the minimum value for the distance d .

RemarkWe let s = d2. So s = (x − 3)2 + y2. Theminimum value for s and the minimum value ford will occur at the same point. (Why?) So wewill work with s instead of d . This approach isfrequently useful in max min distance problems.

Closest Point

Example 2. Problem. Find the point on y = x2 closest to thepoint 〈3, 0〉.

d

xK3 K2 K1 0 1 2 3

y

K1

1

2

3

4

5

6

We let 〈x , y〉 be a point on the curve, and we letd be the distance between 〈x , y〉 and 〈3, 0〉. Sod =

√(x − 3)2 + (y − 0)2 =

√(x − 3)2 + y2.

We want to find the point 〈x , y〉that gives the minimum value for the distance d .

RemarkWe let s = d2. So s = (x − 3)2 + y2.

Theminimum value for s and the minimum value ford will occur at the same point. (Why?) So wewill work with s instead of d . This approach isfrequently useful in max min distance problems.

Closest Point

Example 2. Problem. Find the point on y = x2 closest to thepoint 〈3, 0〉.

d

xK3 K2 K1 0 1 2 3

y

K1

1

2

3

4

5

6

We let 〈x , y〉 be a point on the curve, and we letd be the distance between 〈x , y〉 and 〈3, 0〉. Sod =

√(x − 3)2 + (y − 0)2 =

√(x − 3)2 + y2.

We want to find the point 〈x , y〉that gives the minimum value for the distance d .

RemarkWe let s = d2. So s = (x − 3)2 + y2. Theminimum value for s and the minimum value ford will occur at the same point. (Why?)

So wewill work with s instead of d . This approach isfrequently useful in max min distance problems.

Closest Point

Example 2. Problem. Find the point on y = x2 closest to thepoint 〈3, 0〉.

d

xK3 K2 K1 0 1 2 3

y

K1

1

2

3

4

5

6

We let 〈x , y〉 be a point on the curve, and we letd be the distance between 〈x , y〉 and 〈3, 0〉. Sod =

√(x − 3)2 + (y − 0)2 =

√(x − 3)2 + y2.

We want to find the point 〈x , y〉that gives the minimum value for the distance d .

RemarkWe let s = d2. So s = (x − 3)2 + y2. Theminimum value for s and the minimum value ford will occur at the same point. (Why?) So wewill work with s instead of d .

This approach isfrequently useful in max min distance problems.

Closest Point

Example 2. Problem. Find the point on y = x2 closest to thepoint 〈3, 0〉.

d

xK3 K2 K1 0 1 2 3

y

K1

1

2

3

4

5

6

We let 〈x , y〉 be a point on the curve, and we letd be the distance between 〈x , y〉 and 〈3, 0〉. Sod =

√(x − 3)2 + (y − 0)2 =

√(x − 3)2 + y2.

We want to find the point 〈x , y〉that gives the minimum value for the distance d .

RemarkWe let s = d2. So s = (x − 3)2 + y2. Theminimum value for s and the minimum value ford will occur at the same point. (Why?) So wewill work with s instead of d . This approach isfrequently useful in max min distance problems.

Closest Point

Now s = (x − 3)2 + y2 = (x − 3)2 + x4 since y = x2.

Note that s is a polynomial and so continuous everywhere.As usual, we compute the derivative of s.

Sods

dx= 2(x − 3) + 4x3 = 2(2x3 + x − 3) =

2(x − 1)(2x2 + 2x + 3) = 4(x − 1)(x2 + x + 32) =

4(x − 1)(x2 + 2(12)x + 1

4 + 54) = 4(x − 1)((x + 1

2)2 + 54).

RemarkLet g(x) = 2x3 + x − 3. Note that g(1) = 0. So (x − 1) divides2x3 + x − 3. We find the quotient by long division. Also, we findthat x2 + x + 3

2 = (x + 12)2 + 5

4 by “completing the square”.

Closest Point

Now s = (x − 3)2 + y2 = (x − 3)2 + x4 since y = x2.Note that s is a polynomial and so continuous everywhere.

As usual, we compute the derivative of s.

Sods

dx= 2(x − 3) + 4x3 = 2(2x3 + x − 3) =

2(x − 1)(2x2 + 2x + 3) = 4(x − 1)(x2 + x + 32) =

4(x − 1)(x2 + 2(12)x + 1

4 + 54) = 4(x − 1)((x + 1

2)2 + 54).

RemarkLet g(x) = 2x3 + x − 3. Note that g(1) = 0. So (x − 1) divides2x3 + x − 3. We find the quotient by long division. Also, we findthat x2 + x + 3

2 = (x + 12)2 + 5

4 by “completing the square”.

Closest Point

Now s = (x − 3)2 + y2 = (x − 3)2 + x4 since y = x2.Note that s is a polynomial and so continuous everywhere.As usual, we compute the derivative of s.

Sods

dx= 2(x − 3) + 4x3 = 2(2x3 + x − 3) =

2(x − 1)(2x2 + 2x + 3) = 4(x − 1)(x2 + x + 32) =

4(x − 1)(x2 + 2(12)x + 1

4 + 54) = 4(x − 1)((x + 1

2)2 + 54).

RemarkLet g(x) = 2x3 + x − 3. Note that g(1) = 0. So (x − 1) divides2x3 + x − 3. We find the quotient by long division. Also, we findthat x2 + x + 3

2 = (x + 12)2 + 5

4 by “completing the square”.

Closest Point

Now s = (x − 3)2 + y2 = (x − 3)2 + x4 since y = x2.Note that s is a polynomial and so continuous everywhere.As usual, we compute the derivative of s.

Sods

dx= 2(x − 3) + 4x3 =

2(2x3 + x − 3) =

2(x − 1)(2x2 + 2x + 3) = 4(x − 1)(x2 + x + 32) =

4(x − 1)(x2 + 2(12)x + 1

4 + 54) = 4(x − 1)((x + 1

2)2 + 54).

RemarkLet g(x) = 2x3 + x − 3. Note that g(1) = 0. So (x − 1) divides2x3 + x − 3. We find the quotient by long division. Also, we findthat x2 + x + 3

2 = (x + 12)2 + 5

4 by “completing the square”.

Closest Point

Now s = (x − 3)2 + y2 = (x − 3)2 + x4 since y = x2.Note that s is a polynomial and so continuous everywhere.As usual, we compute the derivative of s.

Sods

dx= 2(x − 3) + 4x3 = 2(2x3 + x − 3) =

2(x − 1)(2x2 + 2x + 3) =

4(x − 1)(x2 + x + 32) =

4(x − 1)(x2 + 2(12)x + 1

4 + 54) = 4(x − 1)((x + 1

2)2 + 54).

RemarkLet g(x) = 2x3 + x − 3. Note that g(1) = 0. So (x − 1) divides2x3 + x − 3. We find the quotient by long division. Also, we findthat x2 + x + 3

2 = (x + 12)2 + 5

4 by “completing the square”.

Closest Point

Now s = (x − 3)2 + y2 = (x − 3)2 + x4 since y = x2.Note that s is a polynomial and so continuous everywhere.As usual, we compute the derivative of s.

Sods

dx= 2(x − 3) + 4x3 = 2(2x3 + x − 3) =

2(x − 1)(2x2 + 2x + 3) = 4(x − 1)(x2 + x + 32) =

4(x − 1)(x2 + 2(12)x + 1

4 + 54) = 4(x − 1)((x + 1

2)2 + 54).

RemarkLet g(x) = 2x3 + x − 3. Note that g(1) = 0. So (x − 1) divides2x3 + x − 3. We find the quotient by long division. Also, we findthat x2 + x + 3

2 = (x + 12)2 + 5

4 by “completing the square”.

Closest Point

Now s = (x − 3)2 + y2 = (x − 3)2 + x4 since y = x2.Note that s is a polynomial and so continuous everywhere.As usual, we compute the derivative of s.

Sods

dx= 2(x − 3) + 4x3 = 2(2x3 + x − 3) =

2(x − 1)(2x2 + 2x + 3) = 4(x − 1)(x2 + x + 32) =

4(x − 1)(x2 + 2(12)x + 1

4 + 54) =

4(x − 1)((x + 12)2 + 5

4).

RemarkLet g(x) = 2x3 + x − 3. Note that g(1) = 0. So (x − 1) divides2x3 + x − 3. We find the quotient by long division. Also, we findthat x2 + x + 3

2 = (x + 12)2 + 5

4 by “completing the square”.

Closest Point

Now s = (x − 3)2 + y2 = (x − 3)2 + x4 since y = x2.Note that s is a polynomial and so continuous everywhere.As usual, we compute the derivative of s.

Sods

dx= 2(x − 3) + 4x3 = 2(2x3 + x − 3) =

2(x − 1)(2x2 + 2x + 3) = 4(x − 1)(x2 + x + 32) =

4(x − 1)(x2 + 2(12)x + 1

4 + 54) = 4(x − 1)((x + 1

2)2 + 54).

RemarkLet g(x) = 2x3 + x − 3. Note that g(1) = 0. So (x − 1) divides2x3 + x − 3. We find the quotient by long division. Also, we findthat x2 + x + 3

2 = (x + 12)2 + 5

4 by “completing the square”.

Closest Point

Now s = (x − 3)2 + y2 = (x − 3)2 + x4 since y = x2.Note that s is a polynomial and so continuous everywhere.As usual, we compute the derivative of s.

Sods

dx= 2(x − 3) + 4x3 = 2(2x3 + x − 3) =

2(x − 1)(2x2 + 2x + 3) = 4(x − 1)(x2 + x + 32) =

4(x − 1)(x2 + 2(12)x + 1

4 + 54) = 4(x − 1)((x + 1

2)2 + 54).

RemarkLet g(x) = 2x3 + x − 3.

Note that g(1) = 0. So (x − 1) divides2x3 + x − 3. We find the quotient by long division. Also, we findthat x2 + x + 3

2 = (x + 12)2 + 5

4 by “completing the square”.

Closest Point

Now s = (x − 3)2 + y2 = (x − 3)2 + x4 since y = x2.Note that s is a polynomial and so continuous everywhere.As usual, we compute the derivative of s.

Sods

dx= 2(x − 3) + 4x3 = 2(2x3 + x − 3) =

2(x − 1)(2x2 + 2x + 3) = 4(x − 1)(x2 + x + 32) =

4(x − 1)(x2 + 2(12)x + 1

4 + 54) = 4(x − 1)((x + 1

2)2 + 54).

RemarkLet g(x) = 2x3 + x − 3. Note that g(1) = 0.

So (x − 1) divides2x3 + x − 3. We find the quotient by long division. Also, we findthat x2 + x + 3

2 = (x + 12)2 + 5

4 by “completing the square”.

Closest Point

Now s = (x − 3)2 + y2 = (x − 3)2 + x4 since y = x2.Note that s is a polynomial and so continuous everywhere.As usual, we compute the derivative of s.

Sods

dx= 2(x − 3) + 4x3 = 2(2x3 + x − 3) =

2(x − 1)(2x2 + 2x + 3) = 4(x − 1)(x2 + x + 32) =

4(x − 1)(x2 + 2(12)x + 1

4 + 54) = 4(x − 1)((x + 1

2)2 + 54).

RemarkLet g(x) = 2x3 + x − 3. Note that g(1) = 0. So (x − 1) divides2x3 + x − 3.

We find the quotient by long division. Also, we findthat x2 + x + 3

2 = (x + 12)2 + 5

4 by “completing the square”.

Closest Point

Now s = (x − 3)2 + y2 = (x − 3)2 + x4 since y = x2.Note that s is a polynomial and so continuous everywhere.As usual, we compute the derivative of s.

Sods

dx= 2(x − 3) + 4x3 = 2(2x3 + x − 3) =

2(x − 1)(2x2 + 2x + 3) = 4(x − 1)(x2 + x + 32) =

4(x − 1)(x2 + 2(12)x + 1

4 + 54) = 4(x − 1)((x + 1

2)2 + 54).

RemarkLet g(x) = 2x3 + x − 3. Note that g(1) = 0. So (x − 1) divides2x3 + x − 3. We find the quotient by long division.

Also, we findthat x2 + x + 3

2 = (x + 12)2 + 5

4 by “completing the square”.

Closest Point

Now s = (x − 3)2 + y2 = (x − 3)2 + x4 since y = x2.Note that s is a polynomial and so continuous everywhere.As usual, we compute the derivative of s.

Sods

dx= 2(x − 3) + 4x3 = 2(2x3 + x − 3) =

2(x − 1)(2x2 + 2x + 3) = 4(x − 1)(x2 + x + 32) =

4(x − 1)(x2 + 2(12)x + 1

4 + 54) = 4(x − 1)((x + 1

2)2 + 54).

RemarkLet g(x) = 2x3 + x − 3. Note that g(1) = 0. So (x − 1) divides2x3 + x − 3. We find the quotient by long division. Also, we findthat x2 + x + 3

2 = (x + 12)2 + 5

4 by “completing the square”.

Closest Point

Now s = (x − 3)2 + y2 = (x − 3)2 + x4 since y = x2.Note that s is a polynomial and so continuous everywhere.As usual, we compute the derivative of s.

Sods

dx= 2(x − 3) + 4x3 = 2(2x3 + x − 3) =

2(x − 1)(2x2 + 2x + 3) = 4(x − 1)(x2 + x + 32) =

4(x − 1)(x2 + 2(12)x + 1

4 + 54) = 4(x − 1)((x + 1

2)2 + 54).

RemarkLet g(x) = 2x3 + x − 3. Note that g(1) = 0. So (x − 1) divides2x3 + x − 3. We find the quotient by long division. Also, we findthat x2 + x + 3

2 = (x + 12)2 + 5

4 by “completing the square”.

Closest Point

(x + 12)2 + 5

4 ≥54 > 0 for all x .

So if x < 1, then x − 1 < 0, and sods

dx= 4(x − 1)((x + 1

2)2 + 54) < 0.

So s is decreasing on (−∞, 1].So if x > 1, then x − 1 > 0, and sods

dx= 4(x − 1)((x + 1

2)2 + 54) > 0.

So s is increasing on [1,∞).So the absolute minimum value for s (and d) occurs with x = 1.So 〈1, 1〉 is the point on y = x2 closest to 〈3, 0〉.

Closest Point

(x + 12)2 + 5

4 ≥54 > 0 for all x .

So if x < 1, then x − 1 < 0,

and sods

dx= 4(x − 1)((x + 1

2)2 + 54) < 0.

So s is decreasing on (−∞, 1].So if x > 1, then x − 1 > 0, and sods

dx= 4(x − 1)((x + 1

2)2 + 54) > 0.

So s is increasing on [1,∞).So the absolute minimum value for s (and d) occurs with x = 1.So 〈1, 1〉 is the point on y = x2 closest to 〈3, 0〉.

Closest Point

(x + 12)2 + 5

4 ≥54 > 0 for all x .

So if x < 1, then x − 1 < 0, and sods

dx= 4(x − 1)((x + 1

2)2 + 54) < 0.

So s is decreasing on (−∞, 1].So if x > 1, then x − 1 > 0, and sods

dx= 4(x − 1)((x + 1

2)2 + 54) > 0.

So s is increasing on [1,∞).So the absolute minimum value for s (and d) occurs with x = 1.So 〈1, 1〉 is the point on y = x2 closest to 〈3, 0〉.

Closest Point

(x + 12)2 + 5

4 ≥54 > 0 for all x .

So if x < 1, then x − 1 < 0, and sods

dx= 4(x − 1)((x + 1

2)2 + 54) < 0.

So s is decreasing on (−∞, 1].

So if x > 1, then x − 1 > 0, and sods

dx= 4(x − 1)((x + 1

2)2 + 54) > 0.

So s is increasing on [1,∞).So the absolute minimum value for s (and d) occurs with x = 1.So 〈1, 1〉 is the point on y = x2 closest to 〈3, 0〉.

Closest Point

(x + 12)2 + 5

4 ≥54 > 0 for all x .

So if x < 1, then x − 1 < 0, and sods

dx= 4(x − 1)((x + 1

2)2 + 54) < 0.

So s is decreasing on (−∞, 1].So if x > 1, then x − 1 > 0,

and sods

dx= 4(x − 1)((x + 1

2)2 + 54) > 0.

So s is increasing on [1,∞).So the absolute minimum value for s (and d) occurs with x = 1.So 〈1, 1〉 is the point on y = x2 closest to 〈3, 0〉.

Closest Point

(x + 12)2 + 5

4 ≥54 > 0 for all x .

So if x < 1, then x − 1 < 0, and sods

dx= 4(x − 1)((x + 1

2)2 + 54) < 0.

So s is decreasing on (−∞, 1].So if x > 1, then x − 1 > 0, and sods

dx= 4(x − 1)((x + 1

2)2 + 54) > 0.

So s is increasing on [1,∞).So the absolute minimum value for s (and d) occurs with x = 1.So 〈1, 1〉 is the point on y = x2 closest to 〈3, 0〉.

Closest Point

(x + 12)2 + 5

4 ≥54 > 0 for all x .

So if x < 1, then x − 1 < 0, and sods

dx= 4(x − 1)((x + 1

2)2 + 54) < 0.

So s is decreasing on (−∞, 1].So if x > 1, then x − 1 > 0, and sods

dx= 4(x − 1)((x + 1

2)2 + 54) > 0.

So s is increasing on [1,∞).

So the absolute minimum value for s (and d) occurs with x = 1.So 〈1, 1〉 is the point on y = x2 closest to 〈3, 0〉.

Closest Point

(x + 12)2 + 5

4 ≥54 > 0 for all x .

So if x < 1, then x − 1 < 0, and sods

dx= 4(x − 1)((x + 1

2)2 + 54) < 0.

So s is decreasing on (−∞, 1].So if x > 1, then x − 1 > 0, and sods

dx= 4(x − 1)((x + 1

2)2 + 54) > 0.

So s is increasing on [1,∞).So the absolute minimum value for s (and d) occurs with x = 1.

So 〈1, 1〉 is the point on y = x2 closest to 〈3, 0〉.

Closest Point

(x + 12)2 + 5

4 ≥54 > 0 for all x .

So if x < 1, then x − 1 < 0, and sods

dx= 4(x − 1)((x + 1

2)2 + 54) < 0.

So s is decreasing on (−∞, 1].So if x > 1, then x − 1 > 0, and sods

dx= 4(x − 1)((x + 1

2)2 + 54) > 0.

So s is increasing on [1,∞).So the absolute minimum value for s (and d) occurs with x = 1.So 〈1, 1〉 is the point on y = x2 closest to 〈3, 0〉.