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A firm’s revenue after selling q items is given by R(q) = 2q3 +36q while the firm’s costs after making q items is C(q) = 3q^2 −10.(a) If the firm can make a maximum of 4 units, how many units should the firm make to maximize profit. What is the profit at this point?(b) If the firm can make a maximum of 10 units, how many units should the firm make to maximize profit?

Question

A firm’s revenue after selling q items is given by R(q) = 2q3 +36q while the firm’s costs after making q items is C(q) = 3q^2 −10.

(a) If the firm can make a maximum of 4 units, how many units should the firm make to maximize profit. What is the profit at this point?

(b) If the firm can make a maximum of 10 units, how many units should the firm make to maximize profit?

check_circleAnswer
Step 1

R(q) = 2q3 +36q

C(q) = 3q2 −10

Profit, P(q) = R(q) - C(q) = 2q3 +36q - (3q2 −10) = 2q3 - 3q2 + 36q+10

If we intend to maximize the profit, then dP/dq = 0 and d2P/dq2 < 0

Step 2

P(q) = 2q3 - 3q2 + 36q+10

Let\\'s differentiate both sides w.r.t q

Recall the famous rule of differentiation: d(xn) / dx = nxn-1

dP / dq = 6q2 - 6q + 36

Setting it to zero, we get 6(q2 - q + 6) = 0

Or, q2 - q + 6 = 0

Discriminant of this quadratic function = (-1)2 - 4 x 1 x 6 = -23 < 0

Hence there is no real root for this equation. Hence, we can\\'t get the point of maximum profit using the usual derivtive route.

Step 3

dP/dq = 6(q2 - q + 6) = 6[(q - 1/2)2 + 23 / 4} > 0

Since dP/dq is always > 0, this means P(q) is continuously in...

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Math

Calculus

Derivative

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