MATH 180 FALL 1995

ASSIGNMENT #31

5.3, #12. y = axe^{−bx}, y′ = a(1 − bx)e^{−bx}, y′′ = ab(bx − 2)e^{−bx}. So y has one critical point, namely x = 1/b. The first-derivative test shows x = 1/b is a local maximum. y′′ has a root at x = 2/b, and y′′ changes sign there. So x = 2/b is an inflection point of y. Changing a in y = axe^{−bx} rescales the graph vertically. Changing b changes the distance 1/b between the origin and the critical point x = 1/b as well as the distance 1/b between the critical point x = 1/b and the inflection point x = 2/b. Indeed, increasing b decreases that distance, decreasing b increases it.

5.3, #13. y = A sin(Bx + C). The parameters A, B, C affect respectively amplitude, period, and phase. See pages 68–69 of the text for examples.

5.4, #2. π(q) = R(q) − C(q) is negative for 0 ≤ q < q_{1}, positive for q_{1} < q < q_{2}, negative for q > q_{2}. In particular, π has a local maximum at a point q = a between q_{1} and q_{2}. C(q) is increasing, so C′(q) ≥ 0 for q ≥ 0. The graph of C(q) appears to flatten out at q = q_{1}, but the slopes get steeper to the right of q_{2}. So C′(q) falls to 0 on [0, q_{1}], and rises for q ≥ q_{1}. The graph of R(q) is a line of positive slope. Thus R′(q) has a horizontal line y = b with b > 0 as its graph. Since π′(a) = R′(a) − C′(a) = 0, the graphs of C′ and R′ intersect at (a, b).

5.4, #5. (a) C(q) = 0.01q^{3} − 0.6q^{2} + 13q is zero when q = 0, so the fixed cost is zero.

(b) The revenue is R(q) = 7q when goods are priced at $7 each. So profit

π(q) = R(q) − C(q) = −0.01q^{3} + 0.6q^{2} − 6q.

π′(q) = −0.03q^{2} + 1.2q − 6 has roots

\frac{−1.2 ± \sqrt{(1.2)^{2} − 4(−6)(−0.03)}}{−0.06} = 5.86, 34.14.

34.14 is a local maximum by the second-derivative test since π′′(q) = −0.06q + 1.2 is negative at q = 34.14. Since goods are sold in discrete quantities, we should round off q to 34 or 35. Now π(34) = $96.56, π(35) = $96.25, So profit is maximized when q = 34.

(c) Suppose q is fixed at 34. Each dollar increase in the price p reduces the quantity sold by 2. Since 34 goods are sold when p = $7, the quantity sold at price p is q = 48 − 2p. We express revenue as the function R(p) = qp = 48p − 2p^{2} of p. The cost is the fixed constant C(34). Thus the profit as a function of p is 48p − 2p^{2} − C(34). This is a parabola opening downwards. The profit is maximal at the vertex. Since π′(p) = 48 − 4p, the critical point p = $12 gives the desired price.