problem of the day no calculator! if f(x) = ln x for 0 < x < 2 x 2 ln 2 for 2 < x < 4 {...
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Problem of the Day No calculator!
If f(x) = ln x for 0 < x < 2x2 ln 2 for 2 < x < 4
{then lim f(x) is x⇒2
A) ln 2 C) ln 16 E) nonexistent B) ln 8 D) 4
Problem of the Day
If f(x) = ln x for 0 < x < 2x2 ln 2 for 2 < x < 4
{then lim f(x) is x⇒2
A) ln 2 C) ln 16 E) nonexistent B) ln 8 D) 4
limit from left does not equal limit from right at 2
Optimization ProblemsOne of the most common uses
of calculus is to determine minimum and maximum values. Some examples aregreatest profit
least costleast timegreatest voltageoptimum sizegreatest strengthgreatest distancegreatest volumeleast materialleast size
A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume?
Method 1 - Guess and check
surface area = area of base + 4 (area of a side)108 = (x)(x) + 4(x)
(h)
xxh
108 = x2 + 4xh
A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume?
Method 1 - Guess and check
xxh
108 = x2 + 4xhvolume = x2h
A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume?
Method 1 - Guess and check
xxh 108 = x2 +
4xhlength of side
volume345678
volume = x2hheight
A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume?
Method 1 - Guess and check
xxh 108 = x2 +
4xh
length of side height
volume345678
74.2592103 ¾108103¼88
8.255.754.1532.111.375
volume = x2h
Method 1 - Guess and check
xxh 108 = x2 +
4xhlength of side height
volume345678
74.2592103 ¾108103¼88
8.255.754.1532.111.375
Any problems with this method?
volume = x2h
A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume?
Method 2 - Use Calculus
xxh
108 = x2 + 4xh
surface area = area of base + 4 (area of a side)
volume = x2h
What is to be maximized?
What is this in Calculus?
A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume?
Method 2 - Use Calculus
xxh
108 = x2 + 4xh
surface area = area of base + 4 (area of a side)
volume = x2h
What is to be maximized? VolumeWhat is this in Calculus? Critical Number
And Critical Numbers come from?
A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume?
Method 2 - Use Calculus
xxh
108 = x2 + 4xh
surface area = area of base + 4 (area of a side)
volume = x2h
What is to be maximized? VolumeWhat is this in Calculus? Critical NumberAnd Critical Numbers come from? First Derivative Set Equal to Zero
A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume?
Method 2 - Use Calculus
xxh
108 = x2 + 4xh
surface area = area of base + 4 (area of a side)
volume = x2h
How can we eliminate a variable?
A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume?
Method 2 - Use Calculus
xxh
108 = x2 + 4xh
surface area = area of base + 4 (area of a side)
volume = x2h
How can we eliminate a variable? Use the equation not being maximized to solve for one of the variables and substitute the result into the equation being maximized.
Procedure to Maximizing or Minimizing1) Develop 2 equations2) Use equation not being maximized to eliminate a variable3) Determine feasible domain for x4) Differentiate5) Find critical numbers6) Evaluate at endpoints of domain and at critical numbers7) Pick value that gives the largest or smallest value
1) Develop 2 equations108 = x2 +
4xhvolume = x2h
2) Use equation not being maximized to eliminate a variable108 = x2 +
4xhh = 108 - x2
4x
V = x2 hV= (x2 ) 108 - x2
4xV = 27x - ¼x3
3) Determine feasible domain for x1) x must be greater than zero
2) The largest x would be when the height is zero. Then 108 = x2 + 0 and x2 = √108
Thus 0 < x < √1084) DifferentiateV = 27x -
¼x3
dV = 27 - ¾x2
dx
dV = 27 - ¾x2
dx
0 = 27 - ¾x2
3x2 = 108 x = ±6
5) Find critical numbers
6) Evaluate at endpoints of domain and at critical numbersf(0) = 0
f(6) = 108f(√108) = 0
f(0) = 0f(6) = 108f(√108) = 0
7) Pick value that gives the largest or smallest value
maximum
so x = 6 and h = 3
dimensions are 6 x 6 x 3
Find 2 positive integers whose product is 192 and the sum of the first plus three times the second is a minimum.
Find 2 positive integers whose product is 192 and the sum of the first plus three times the second is a minimum.1) Develop 2 equations
2) Use equation not being maximized to eliminate a variable
Find 2 positive integers whose product is 192 and the sum of the first plus three times the second is a minimum.3) Determine feasible domain for x
4) Differentiate
5) Find critical numbers
Find 2 positive integers whose product is 192 and the sum of the first plus three times the second is a minimum.6) Evaluate at endpoints of domain and at critical numbers
7) Pick value that gives the largest or smallest value
Find 2 positive integers whose product is 192 and the sum of the first plus three times the second is a minimum.
The numbers are 24 and 8
Find the length and width of a rectangle that has an area of 64 and a minimum perimeter.
Find the length and width of a rectangle that has an area of 64 and a minimum perimeter.
Area = l . w64 = l . w64/w = l
Perimeter = 2l + 2wPerimeter = 2(64/w) + 2w
y' = -128/w2 + 20 = -128/w2 + 28 = wLength = 8 and
width = 8
Find the point on the graph of the function f(x) = x2 that is closest to the point (2, ½).
Find the point on the graph of the function f(x) = x2 that is closest to the point (2, ½).
y = x2
distance = √(2 - x)2 + (½ - y)2
distance = (x4 - 4x + 4.25)½
distance ' = ½(x4 - 4x + 4.25)-½ (4x3 - 4) 0 = ½(x4 - 4x + 4.25)-½ (4x3 -
4)1 = x
A tank with a rectangular base and rectangular sides is to be open at the top. It is to be constructed so that its width is 4 meters and its volume is 36 cubic meters. If building the tank costs $10 per square meter for the base and $5 per square meter for the sides, what is the cost of the least expensive tank?
A tank with a rectangular base and rectangular sides is to be open at the top. It is to be constructed so that its width is 4 meters and its volume is 36 cubic meters. If building the tank costs $10 per square meter for the base and $5 per square meter for the sides, what is the cost of the least expensive tank?
$330
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