MA 1505 Mathematics ITutorial 6 Solutions

1. V = I ×R =⇒ I = VR .

(i)∂I

∂V=

1R

.

If R = 15, then∂I

∂V=

115≈ 0.0667 A/V.

(ii)∂I

∂R= − V

R2. If V = 120 and R = 20, then

∂I

∂R= − 120

202= −0.3 A/Ω.

(iii) By Chain rule,dI

dt=

∂I

∂V

dV

dt+∂I

∂R

dR

dt=

1R

dV

dt− V

R2

dR

dt.

Since R = 400, I = 0.08, V = 32,dV

dt= −0.01,

dR

dt= 0.03, so

dI

dt=

1400

(−0.01)− (0.08)(400)4002

(0.03) = −3.1× 10−5.

2. fx = e2y−x + xe2y−x(−1) = e2y−x(1− x) and fy = 2xe2y−x.

So fx(−2,−1) = 3 and fy(−2,−1) = −4.

(i) u =1√2i +

1√2j is a unit vector. Thus,

Duf(−2,−1) = 3× 1√2− 4× 1√

2= − 1√

2= −√

22.

(ii) The unit vector in the direction of 3i + 4j is u =35i +

45j. Thus,

Duf(−2,−1) = 3× 35− 4× 4

5= − 7

5.

Let u = ai + bj be a unit vector. Then

Duf(−2,−1) = fx(−2,−1)× a+ fy(−2,−1)× b= (fx(−2,−1)i + fy(−2,−1)j) • (ai + bj)

= ‖fx(−2,−1)i + fy(−2,−1)j‖ ‖u‖ cos θ (∗)

where θ is the angle between fx(−2,−1)i + fy(−2,−1)j and u.

Since the largest value of cos θ is 1 which occurs when θ = 0, this means that the largestpossible value of Duf(−2,−1) occurs when u is in the same direction as fx(−2,−1)i +fy(−2,−1)j = 3i− 4j.

MA1505 Tutorial 6 Solutions

3. fx = yz cos (xyz), fy = xz cos (xyz) and fz = xy cos (xyz).

So fx(12,13, π) =

√3

6π, fy(

12,13, π) =

√3

4π and fz(

12,13, π) =

√3

12.

(i) Let u =1√3i− 1√

3j +

1√3k.

Thus, the rate of change of f at P in the direction u is given by

Duf

(12,13, π

)=√

36π × 1√

3+√

34π ×

(− 1√

3

)+√

312× 1√

3=

112

(1− π).

(ii) The change in the value of f :

df ≈ Duf

(12,13, π

)× dt =

112

(1− π)× 0.1 ≈ −0.01785.

So the value of f will decrease by about 0.01785 unit.

4. (i) Let f(x, y) = ln (x2y)− xy − 2x = 2 lnx+ ln y − xy − 2x, where x > 0, y > 0.

We have fx =2x− y− 2 and fy =

1y− x. Also, fxx = − 2

x2, fxy = −1 and fyy = − 1

y2. Then

fy = 0 implies that x =1y

, and substitution into fx = 0 gives 2y − y − 2 = 0, i.e. y = 2. So

x = 1/2. Thus, the only critical point is (1/2, 2).

Now D(1/2, 2) = (−8)(−1/4) − 12 = 1 > 0, fxx(1/2, 2) = −8 < 0 implies that f(1/2, 2) =− ln 2− 2 is a local maximum.

(ii) Let g(x, y) = xy(1− x− y).

We have gx = y − 2xy − y2, gy = x− x2 − 2xy, gxx = −2y, gyy = −2x and gxy = 1− 2x− 2y.

Then gx = 0 implies y = 0 or y = 1−2x. Substituting y = 0 into gy = 0 gives x = 0 or x = 1.Substituting y = 1− 2x into gy = 0 gives x = 0 or x = 1/3.

Thus the critical points are (0, 0), (1, 0), (0, 1) and (1/3, 1/3).

Now D(0, 0) = D(1, 0) = D(0, 1) = −1 while D(1/3, 1/3) = 1/3 and gxx(1/3, 1/3) = −2/3 <0.

Thus (0, 0), (1, 0) and (0, 1) are saddle points, and g(1/3, 1/3) = 1/27 is a local maximum.

(iii) Let h(x, y) = x2 + y2 + x−2y−2.

We have hx = 2x− 2x−3y−2, hy = 2y− 2x−2y−3, hxx = 2 + 6x−4y−2, hyy = 2 + 6x−2y−4 andhxy = 4x−3y−3.

Then hx = 0 implies that 2x4y2 − 2 = 0 or y2 = x−4. Note that neither x nor y can be zero.Now hy = 0 implies that 2x2y4 − 2 = 0, and with y2 = x−4 this implies 2x−6 − 2 = 0 orx6 = 1, that is x = ±1. If x = 1, then y = ±1. If x = −1, then y = ±1.

So the critical points are (1, 1), (1,−1), (−1, 1) and (−1,−1).

Now D(±1,±1) = D(±1,∓1) = 64 − 16 > 0 and hxx is always greater than zero. Thush(±1,±1) = h(±1,∓1) = 3 are local minima.

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MA1505 Tutorial 6 Solutions

5. Let v =∂u

∂y, then the given condition u

∂2u

∂x∂y=∂u

∂x

∂u

∂yimplies that u

∂v

∂x=∂u

∂xv.

Therefore we haveu∂v

∂x− v∂u

∂xu2

= 0,

and hence∂

∂x( v

u) = 0. This means that vu is a function of one variable y only. Now the result

follows by observing that vu = 1

u

∂u

∂y=∂(lnu)∂y

.

6. The function P (L,K) is maximized, subject to the constraint L+K = 150.

LetP (L,K, λ) = 50L2/5K3/5 − λ(L+K − 150).

Then set

∂P

∂L= 20L−3/5K3/5 − λ = 0

∂P

∂K= 30L2/5K−2/5 − λ = 0

∂P

∂λ= −L−K + 150 = 0.

Eliminate λ from the first two equations to get

20L−3/5K3/5 = 30L2/5K−2/5

K =32L.

Substituting this into the third equation yields

L = 60 and K = 90.

Thus, P (60, 90) = 50(60)2/5(90)3/5 ≈ 3826.

Answer: Allocate $60,000 to labour and $90,000 to capital investment to generate a maximumoutput of approximately 3,826 units.

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