economics

Solution to Maximization Problems October 30, 2020

1. [Revenue Maximization] Maximize the following total revenue function (TR) and the profit π by finding out (a) the critical values (the output values); (b) checking the second order conditions and (c) the maximum values. Note I: First Order Condition (F.O.C.) The critical values are the set of points in the domain the derivative of the total revenue or profit function is equal to zero. Total Revenue

T R = 32Q − Q2.

(a) F. O. C. dT R dQ

= 32 − 2Q = 0,

yields the critical point (the output Q where the F.O.C. is satisfied) as

Q∗ = 32 2

= 16.

(b) The maximum total revenue is

T R∗ = T R(Q∗) = 32 × 16 − 162 = 512 − 256 = 256.

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Solution to Maximization Problems October 30, 2020

2. [Profit Maximization] Total Profit Function (π)

π = − Q3

3 − 5Q2 + 2000Q − 326.

(a) F. O. C. dπ dQ

= − 3Q2

3 − 5 ×(2Q)+ 2000 = 0,

which simplifies to −Q2 − 10Q + 2000 = 0,

and can be factored as

−Q2 − 50Q + 40Q + 2000 = −Q(Q + 50)+ 40(Q + 50) = (Q + 50)(−Q + 40) = 0,

which yields the critical point as Q∗ = 40.

Observe that Q = −50 also satisfies the F. O. C. but is not meaningful and hence we ignore it. (b) The maximum profit is

π∗ = π(Q∗) = − 403

3 −5×402 +2000×40−326 = −

64000 3

−8000+80000−326 = 50340.67.

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Solution to Maximization Problems October 30, 2020

3. [Minimum Average Cost] Minimize the average cost for each of the following total cost function (TC) by finding out (a) average cost function, (b) the critical values (the output values) at which AC is minimized; and (c) the minimum average cost. Total Cost Function

T C = Q3 − 21Q2 + 500Q.

(a) The average cost is

AC = T C Q

= Q3 − 21Q2 + 500Q

Q = Q2 − 21Q + 500.

F. O. C. dAC dQ

= 2Q − 21 = 0,

which yields the critical point as

Q∗ = 10.5.

(b) The minimum AC is

AC∗ = AC(Q∗) = (10.5)2 − 21 × 10.5 + 500 = 110.25 − 220.5 + 500 = 389.75.

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Solution to Maximization Problems October 30, 2020

4. [Profit Maximization - II] Given the following total revenue function (TR) and the total cost function (TC), maximize profit π by following steps (a) set up the profit function

π = T R − T C,

(b) the critical values (the output values) where the profit is at a relative extremum; (c) checking the second order conditions and (d) the maximum profit value.

T R = 4350Q − 13Q2;

T C = Q3 − 5.5Q2 + 150Q + 675.

(a) The profit is

π = 4350Q − 13Q2 − Q3 + 5.5Q2 − 150Q − 675 = 4200Q − 7.5Q2 − Q3 − 675.

F. O. C. dπ dQ

= 4200 − 15Q − 3Q2 = 0,

which can be factored as

4200 − 120Q + 105Q − 3Q2 = 120(35 − Q)+ 3Q(35 − Q) = (35 − Q)(105 + 3Q) = 0.

This yields the critical point as Q∗ = 35,

as the other solution Q = −40 is not economically meaningful. (b) The maximum profit is

π∗ = π(Q∗) = 4200 × 35 − 7.5(35)2 −(35)3 − 675 = 94262.5.

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