Lab9 Graded Solutions

ECONOMICS 207FALL 2015

HOMEWORK LAB 9

Problem 1. In the following problems you are given a production function for a firm wherey is the level of output and x is the level of the variable input. You are given the price (p)ofthe output and the price (w) of the single variable input. For each problem write down anequation that represents profit for the firm. Then find the optimal level of input, of output,of cost and of profit.

a.output price = p = 20

input price = w = 500

y = output = f(x) = 100x + 20x2 − x3

Profit = py − wx = p(100x+ 20x2 − x3)− wx

First-order condition:

dProfit

dx= p(100 + 40x− 3x2)− w = 0 ⇐⇒ 100− w/p+ 40x− 3x2 = 0

⇐⇒ 75 + 40x− 3x2 = 0⇒ x = 0 or 15

Second order conditions:

d2Profit

dx2= p(40− 6x);

d2Profit

dx2(15)− 1000 < 0

so x = 15 is a profit maximizing input level.

f(15) = 2625

cost = wx = 7500

profit = 20 · 2625− 7500 = 45000

2 ECONOMICS 207 FALL 2015 HOMEWORK LAB 9

b.output price = p = 4

input price = w = 6400

y = output = f(x) = 400x + 100x2 − x3

Profit = py − wx = p(400x+ 100x2 − x3)− wx

First-order condition:

dProfit

dx= p(400 + 200x− 3x2)− w = 0 ⇐⇒ 400− w/p+ 200x− 3x2 = 0

⇐⇒ −1200 + 200x− 3x2 = 0⇒ x = 20/3 or 60

Second order conditions:

d2Profit

dx2= p(200− 6x);

d2Profit

dx2(20/3) = 640 > 0;

d2Profit

dx2(60) = −640 < 0

so x = 60 is the profit maximizing input level.

f(60) = 168000

cost = wx = 384000

profit = 4 · 168000− 384000 = 288000

ECONOMICS 207 FALL 2015 HOMEWORK LAB 9 3

Problem 2. For each of the following problems you are given a price (p) and the costfunction for a competitive firm. TC stands for total cost and y represents the level of output.Find the profit maximizing level of output for each case.

a.price = p = 500

Total Cost = TC = 500 + 125y − 10y2 + y3

Profit = TR− TC = py − TC.First-order condition:

dProfit

dy= p− 125 + 20y − 3y2 = 0 ⇐⇒ 375 + 20y − 3y2 = 0

⇒ y = 15 or 0

Second order condition:

d2Profit

dy2= 20− 6y;

d2Profit

dy2(15) = −70 < 0

so y = 15 is the profit maximizing output level.


b.p = 600

TC = 600 + 200y − 20y2 + y3


dProfit

dy= p− 200 + 40y − 3y2 = 0 ⇐⇒ 400 + 40y − 3y2 = 0

⇒ y = 20 or 0


d2Profit

dy2= 40− 6y;

d2Profit

dy2(20) = −40 < 0



c.p = 596

TC = 800 + 500y − 50y2 + 3y3


dProfit

dy= p− 500 + 100y − 9y2 = 0 ⇐⇒ 96 + 100y − 9y2 = 0

⇒ y = 12 or 0


d2Profit

dy2= 100− 18y;

d2Profit

dy2(12) = −20 < 0



Problem 3. Find the first and all second order partial derivatives of each of the followingfunctions. Write down the Hessian matrix in each case and obtain its determinant.

a. z = 30x1/4y3/4 − 6x − 3y

∂z

∂x= (30/4)x−3/4y3/4 − 6

∂z

∂y= (90/4)x1/4y−1/4 − 3

∂2z

∂x2= −(90/16)x−7/4y3/4

∂2z

∂y2= −(90/16)x1/4y−5/4

∂2z

∂x∂y=

∂2z

∂y∂x= (90/16)x−3/4y−1/4

H =

[−(90/16)x−7/4y3/4 (90/16)x−3/4y−1/4

(90/16)x−3/4y−1/4 −(90/16)x1/4y−5/4

]det(H) = (−90/16)2x−6/4y−2/4 −

((90/16)x−3/4y−1/4

)2= 0


b. y = 10x1/21 x

1/32 + 2x21x2 − 3x2x

31 + 4x1x2 − 5x3x

21 + 3x3

∂y

∂x1= (10/2)x

−1/21 x

1/32 + 4x1x2 − 9x21x2 + 4x2 − 10x3x1

∂y

∂x2= (10/3)x

1/21 x

−2/32 + 2x21 − 3x31 + 4x1

∂y

∂x3= −5x21 + 3

∂2y

∂x21= −(10/4)x−3/2

1 x1/32 + 4x2 − 18x1x2 − 10x3

∂2y

∂x22= −(20/9)x1/21 x

−5/32

∂2y

∂x23= 0

∂2y

∂x1∂x2= (10/6)x

−1/21 x

−2/32 + 4− 9x21 + 4

∂2y

∂x1∂x3= −10x1

∂2y

∂x2∂x3= 0

H =

−(10/4)x−3/21 x

1/32 + 4x2 − 18x1x2 − 10x3 (10/6)x

−1/21 x

−2/32 + 4x1 − 9x21 + 4 −10x1

(10/6)x−1/21 x

−2/32 + 4− 9x21 + 4 −(20/9)x1/21 x

−5/32 0

−10x1 0 0

We use the third row to make finding the determinant easier

det(H) = −10x1(0−

(−10x1

(−(20/9)x1/21 x

−5/32

)))− 0(. . . ) + 0(. . . )

= −(10x1)2(−(20/9)x1/21 x

−5/32

)

Lab9 Graded Solutions

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