Optimization, question 9

Topic: Optimization
Difficulty: Standard
Suggested time: 5 min
Marks: 3
Theory: UBC Math 100 §8 CLP-1 §3.5Key results

A furniture workshop’s total cost of making $q$ chairs a month is $C(q)$ dollars, given below. Find the smallest possible average cost per chair.

C(q) = 2560 + 25q + 0.1q^{2}

Give the answer in dollars per chair.

Your answer

A whole number.

How sure are you?

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Theory UBC Math 100 §8 CLP-1 §3.5Key results

Worked solution

Strategy

Average cost is C(q)/q. Minimise it over q > 0 like any one-variable function, justify the minimum, and use the fact that average cost equals marginal cost at the optimum as a check.

Working

$A(q) = \frac{C(q)}{q} = \frac{2560}{q} + 25 + 0.1q, \qquad q > 0$
Divide every term by q; the fixed cost is spread over more chairs as output grows.
$A'(q) = -\frac{2560}{q^{2}} + 0.1 = 0 \quad\Longrightarrow\quad q^{2} = 25600 \quad\Longrightarrow\quad q = 160$
Only the positive root is an output.
$A''(q) = \frac{5120}{q^{3}} > 0 \ \text{for } q > 0$
A is concave up on its whole domain, so q = 160 is the global minimum.
$A(160) = 16 + 25 + 16 = 57, \qquad C'(160) = 25 + 0.2(160) = 57$
Average cost equals marginal cost at the minimum, as the next line shows it must.
$A'(q) = \frac{q\,C'(q) - C(q)}{q^{2}} = 0 \iff C'(q) = \frac{C(q)}{q}$
The quotient rule: the average stops falling exactly when one more chair costs as much as the average.

Answer

A_{\min} = A(160) = \$57 \text{ per chair}

Check

At 150 chairs the average cost is about $57.07 and at 170 about $57.06, both a little above $57.

Where this goes wrong

Answering 160, the output, when the question asks for the cost, or giving the total cost C(160) = 9120 instead of the cost per chair. The function to minimise is the average C(q)/q, not the total C(q), which only rises with q.

Leans on

The power rule with negative exponents, from Differentiation rules

Try a similar question

Find the minimum value of

g(x) = x + \dfrac{9}{x}

for

x > 0

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Optimization practice question 9