QUIZ #10: CALCULUS 1A (Stankova) Wednesday, April 7, 2004 Section 10:00–11:00 (Voight)

Problem 1. Find two positive numbers whose product is 4 and for which the sum of their squares is a minimum.

Solution. Let x, y be the numbers. Then xy = 4, so y = 4/x And the sum of their squares is

S(x) = x^{2} + y^{2} = x^{2} + (\frac{4}{x})^{2} = x^{2} + \frac{16}{x^{2}};

we want this to be a maximum, so we find

S′(x) = 2x − \frac{32}{x^{3}} = 0

or 2x^{4} − 32 = 2(x^{4} − 16) = 2(x^{2} − 4)(x^{2} + 4) = 2(x − 2)(x + 2)(x^{2} + 4) = 0, so x = 2. Therefore y = 2 as well.

Finally, we verify that this is a local minimum:

S′′(x) = 2 + \frac{96}{x^{4}} > 0.

QUIZ #10: CALCULUS 1A (Stankova) Wednesday, April 7, 2004 Section 11:00–12:00 (Voight)

Problem 1. Consider the cost function

C(x) = 16 + x^{3}.

given in millions of dollars.

(a) At what production level x is the average cost minimal?

(b) Let

p(x) = 36 − 3x − x^{2}

be the demand function. At what production level x is the profit maximal? What is the maximum profit?

Solution. For (a), the average cost is

\frac{C(x)}{x} = \frac{16}{x} + x^{2}

so to minimize we compute the derivative

(\frac{C(x)}{x})′ = \frac{−16}{x^{2}} + 2x = 0

so x^{3} = 8, or x = 2. This is indeed a minimum, as we compute that

(\frac{C(x)}{x})′′ = \frac{32}{x^{3}} + 2 > 0.

For (b), the profit is the revenue minus cost, i.e.

P (x) = xp(x) − C(x) = 36x − 3x^{2} − x^{3} − (16 + x^{3}) = −16 + 36x − 3x^{2} − 2x^{3}

so the profit is maximized when

P′(x) = 36 − 6x − 6x^{2} = −6(x^{2} + x − 6) = −6(x − 2)(x + 3) = 0

i.e. x = 2. This is indeed a local maximum, since

P′′(x) = −6 − 12x < 0

The maximum profit is

P (x) = −16 + 72 − 12 − 16 = 28

million dollars.