(a) Given the total cost function: C(q)=4000+50q+0.002q2C(q)=4000+50q+0.002q2 we can find the average cost function by dividing by the number of units q as follows:

¯C(q)=C(q)q=4000q+50+0.002qC¯(q)=C(q)q=4000q+50+0.002q

Differentiating the average cost function gives:

d¯Cdq=−4000q2+0.002dC¯dq=−4000q2+0.002

The average cost is a minimum when the derivative is zero, therefore:

0=−4000q2+0.0020=−4000q2+0.002

q2=40000.002=2,000,000q2=40000.002=2,000,000

The solution to the above is q≈1414.2136q≈1414.2136 or q≈−1414.2136q≈−1414.2136. Because the units sold are a positive quantity, we can dismiss the negative solution. The average cost is a minimum when 1,414 units are sold.

 

(b) We are given the demand function: p=80−0.025qp=80−0.025q so we can write the expression for revenue generated as:

R(q)=qp=q(80−0.025q)=80q−0.025q2R(q)=qp=q(80−0.025q)=80q−0.025q2

Differentiating the revenue function and finding the value of q which makes the derivative zero gives:

R′(q)=dRdq=80−0.050qR′(q)=dRdq=80−0.050q

0=80−0.050q0=80−0.050q

q=800.050=1600q=800.050=1600

The maximum revenue occurs when q=1600q=1600 units and is given by:

R(1600)=80(1600)−0.025(16002)=64,000R(1600)=80(1600)−0.025(16002)=64,000

The maximum revenue is $64,000.

 

(c) The expression for profit is found by subtracting cost from revenue.

P(q)=80q−0.025q2−(4000+50q+0.002q2)P(q)=80q−0.025q2−(4000+50q+0.002q2)

P(q)=30q−0.027q2−4000P(q)=30q−0.027q2−4000

Differentiating the profit function and finding the value of q which makes the derivative zero gives:

P′(q)=dPdq=30−0.054qP′(q)=dPdq=30−0.054q

0=30−0.054q0=30−0.054q

q=300.054≈555.5556q=300.054≈555.5556

The maximum profit generated is:

P(q)=30(555.5556)−0.027(555.55562)−4000≈4333.33P(q)=30(555.5556)−0.027(555.55562)−4000≈4333.33

The maximum profit is $4,333 rounded to the nearest dollar when 556 units are sold.