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Study Guides > Business Calculus

Reading: Applied Optimization, Part II

“Average Cost = Marginal Cost”

“Average cost is minimized when average cost = marginal cost” is another saying that isn’t quite true; in this case, the correct statement is Average Cost has critical points when Average Cost and Marginal Cost are equal.                Let’s look at a geometric argument here:
Figure 1 Figure 1
Remember that the average cost is the slope of the diagonal line, the line from the origin to the point on the total cost curve. If you move your clear plastic ruler around, you’ll see (and feel) that the slope of the diagonal line is smallest when the diagonal line just touches the cost curve – when the diagonal line is actually a tangent line—when the average cost is equal to the marginal cost.

Example

The cost in dollars to produce q jars of gourmet salsa is given by [latex] C(q) = 160 + 2q + 0.1q^2 [/latex]. Find the quantity where the average cost is minimum.

Solution

[latex] A(q) = \frac{C(q)}{q} = \frac{160}{q} + 2 + 0.1q [/latex]. We could find the critical points by finding A′, or by setting average cost to marginal cost; I’ll do the latter this time. [latex] MC(q) = 2 + 0.2q [/latex]. So I want to solve: [latex-display] \frac{160}{q} + 2 + 0.1q = 2 + 0.2q [/latex-display] [latex-display] \frac{160}{q} = 0.1q [/latex-display] [latex-display] 1600 = q^2 [/latex-display] [latex-display] q = 40 [/latex-display] The critical point of average cost is when q = 40. Notice that we still have to confirm that the critical point is a minimum. For this, we can use the first or second derivative test on A(q). [latex-display] A \prime(q) = \frac{-160}{q^2} + 0.1 [/latex-display] [latex-display] A \prime \prime (q) = \frac{320}{q^3} > 0 [/latex-display] The second derivative is positive for all positive q, so that means this is a local min. Average cost is minimized when they produce 40 jars of salsa; at that quantity, the average cost is $10 per jar. (Mighty expensive salsa.)

Tangent Line Approximation (TLA)

Back when we first thought about the derivative, we used the slope of secant lines over tiny intervals to approximate the derivative: [latex-display] f\prime(a) \cong \frac{\Delta y}{\Delta x} = \frac{f(x) - f(a)}{x -a} [/latex-display] Now that we have other ways to find derivatives, we can exploit this approximation to go the other way. Solve the expression above for f(x), and you’ll get the tangent line approximation: To approximate the value of f(x) using TLA, find some a where
  1. a and x are “close,” and
  2. You know the exact values of both f (a) and f ′(a). Then [latex] [/latex] Another way to look at the same formula: How close is close? It depends on the shape of the graph of f. In general, the closer the better. 

Example

Suppose we know that g(20) = 5 and g’(20) = 1.4. Using just this information, we can approximate the values of g at some nearby points: g(23) ≈ 5 + (1.4)(23 – 20) = 9.2 g(18) ≈ 5 + (1.4)(18 – 20) = 2.2 Note that we don’t know if these approximations are close – but they’re the best we can do with the limited information we have to start with. Note also that 18 and 23 are sort of close to 20, so we can hope these approximations are pretty good. We’d feel more confident using this information to approximate g(20.003). We’d feel very unsure using this information to approximate g(55).