REVIEW CHAPTER 11: SOLUTIONS

Problem 1:

1. The equation of the tangent plane to z = f(x, y) at (1, 2) is given by

z − f(1, 2) = fx(1, 2)(x− 1) + fy(1, 2)(y − 2).

One gets {fx(x, y) = 18x+ 6fy(x, y) = 2y − 3

so fx(1, 2) = 24fy(1, 2) = 1f(1, 2) = 18

Finally, an equation for the tangent plane is

z − 18 = 24(x− 1) + (y − 2).

2. The linearization of f at a, b) is given by

L(x, y) = fx(a, b)(x− a) + fy(a, b)(y − b) + f(a, b).

In our case (a, b) = (1, 2) and we get

L(x, y) = 24x+ y − 8,

sof(0.9, 2.1) ∼= L(0.9, 2.1) = 15.7.

Here is the graph of f and the one of L.

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Figure 1. The graph of f and its linearization.

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2 REVIEW CHAPTER 11: SOLUTIONS

The second picture shows that when (x, y) is close enough to (1, 2), the approx-imation of f(x, y) by L(x, y) is quite good.

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Figure 2. Linearization around (1, 2)

3. Remind the formula

Duf(2, 1) = ∇f(2, 1).u,

where u = a|a| is the unit vector with the same direction as a.

In particular, one gets u = 15 〈−3, 4〉 and

Duf(2, 1) = 〈24, 1〉.15〈−3, 4〉 = −68

5.

4.

∇f(x, y) = 0 ⇔{

18x+ 6 = 02y − 3 = 0

⇔{x = −3y = 3

2

Moreover, f(−3, 32

)= 74.75. So the unique critical point is Q =

(−3, 32 , 74.75

).

5. A calculation gives D(x, y) = 18.2− 0 = 36 > 0 and fxx = 18 > 0 so the criticalpoint corresponds to a local minimum.

Problem 4:

1. We look for the minimum of the function f(x, y, z) = x2 + y2 + z2 (the squareof the distance to the origin) under the constraint g(x, y, z) = xy − z2 + 4 = 0.

The Lagrange multiplier condition tells you we have to solve{∇f(x, y, z) = λ∇g(x, y, z)g(x, y, z) = 0

which gives

REVIEW CHAPTER 11: SOLUTIONS 3

2x = λy (1)2y = λx (2)2z = −2λz (3)xy − z2 + 4 = 0 (4)

First note that λ 6= 0. In fact, if λ = 0, then x = y = z = 0 and equation (4)cannot be satisfied.

If z = 0, then x 6= 0 (because of equation (4)). Equation (1) and (2) gives you4x = λ2x so λ = ±2. Combining equation (1) and (4) gives you λy2 = 8 so λ hasto be positive: λ = 2. So we finally gets x = y from equation (1) or (2) but incontradicts equation (4).

It follows that z 6= 0 and equation (3) gives you λ = −1. So we get x = y = 0from equation (1) and (2) so z = ±2 from equation (4).

Finally, one easily checks that f(0, 0, 2) = f(0, 0,−2) = 4 so both points areglobal minima of f under the constraint g(x, y, z) = 0.

The equation g(x, y, z) = 0 defines a hyperboloid of one-sheet. The followingpicture shows this surface around the origin and we show how the distance is real-ized.

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Figure 3. Realization of the distance to the origin.

2. The surface is a level surface on g so ∇g(1,−3, 1) is normal to the surface atP = (1,−3, 1). One gets

∇g(1,−3, 1) = 〈−3, 1,−2〉.Thus an equation for the tangent plane is

−3(x− 1) + (y + 3)− 2(z − 1) = 0

and a parametric equation for the normal line is x = 1− 3ty = −3 + tz = 1− 2t

4 REVIEW CHAPTER 11: SOLUTIONS

Problem 6: We have f(x, y, z) = xy − z2 and x = r cos ty = cos2 tz = r

So, applying chain rule, we get

∂f

∂r(r cos θ, cos2 θ, r) =

∂f

∂x

∂x

∂r+∂f

∂y

∂y

∂r+∂f

∂z

∂z

∂r

= cos2 θ cos θ + 0− 2r

and

∂f

∂θ(r cos θ, cos2 θ, r) =

∂f

∂x

∂x

∂θ+∂f

∂y

∂y

∂θ+∂f

∂z

∂z

∂θ

= cos2 θ(−r sin θ) + r cos θ(−2 cos θ sin θ) + 0

= −3r cos2 θ sin θ

Problem 8:According to the theory, the absolute extremas of f on the region can be ei-

ther at a critical point inside the region or on the boundary. So we will comparethe maximum (respectively the minimum) of f at the critical points and on theboundary.

Critical points:

∇f(x, y) = 0 ⇔{

3x2 + 2x = 04y = 0

⇔ (x, y) = (0, 0) or (x, y) = (−2/3, 0).

Note that only (0, 0) is in the domain of f . Moreover,

(1) f(0, 0) = 0.

Value of f of the boundary:The boundary of the domain of f consists of 3 segments

∆1 = {(x, 0), x ∈ [0, 1]}∆2 = {(0, y), y ∈ [0, 1]}∆3 = {(x, 1− x), x ∈ [0, 1]}.

The restriction of f on ∆1 is given by

g1(x) = f(x, 0) = x3 + x2.

One easily checks that g1 is increasing on [0, 1] so its extremas are

(2) g1(1) = f(1, 0) = 2 and g1(0) = f(0, 0) = 0.

The restriction of f on ∆2 is given by

g2(y) = f(0, y) = 2y2

so its extremas on [0, 1] are

(3) g2(0) = f(0, 0) = 0 and g2(1) = f(0, 1) = 2.

REVIEW CHAPTER 11: SOLUTIONS 5

The restriction of f on ∆3 is given by

g3(x) = f(x, 1− x) = x3 + 3x2 − 4x+ 2.

On gets g′3(x) = 3x2 = 6x − 4 which is a quadratic polynomial of strictly positive

determinant. Its only root in [0, 1] is xc =√

73 − 1, which is the only critical point

of g3 on [0, 1]. So the extremas are

(4) g3(0) = g3(1) = 2, and g3(xc) = f(xc, 1− xc) ∼= 0.87

Finally, one gets that the global minimum is 0 and the global maximum 2.

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Figure 4. The graph of f .