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?

Step 1

R(q) = 2q^{3} +36q

C(q) = 3q^{2} −10

Profit, P(q) = R(q) - C(q) = 2q^{3} +36q - (3q^{2} −10) = 2q^{3} - 3q^{2} + 36q+10

If we intend to maximize the profit, then dP/dq = 0 and d^{2}P/dq^{2} < 0

Step 2

P(q) = 2q^{3} - 3q^{2} + 36q+10

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

Recall the famous rule of differentiation: d(x^{n}) / dx = nx^{n-1}

dP / dq = 6q^{2} - 6q + 36

Setting it to zero, we get 6(q^{2} - q + 6) = 0

Or, q^{2} - 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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