MATH 180 FALL 1995

ASSIGNMENT #34

5.6, #1. The fencing forms 3 sides of a rectangle with the wall forming the 4th side. Let the 3 sides have lengths x, y, x, so that 2x + y = 100. The rectangle then has area A = xy = x(100 − 2x) = 100x − 2x^{2}. The problem: Maximize A for 0 ≤ x < ∞. Since A is a parabola opening downwards, the maximum occurs at the vertex. A′ = 100 − 4x is zero at x = 25. The largest area that can be enclosed is A(25) = 1250 square feet.

5.6, #3. Let the bottom of the box be a square of side length x, and let h be the height of the box. The box has volume V = hx^{2} and surface area S = x^{2} + 4hx (one bottom area and 4 side areas). So S = x^{2} + (4V/x). The problem: Minimize S for 0 < x < ∞. Now S′ = 2x − (4V/x^{2}) = (2x^{3} − 4V)/x^{2} is zero at x = (2V)^{1/3}. Moreover, S′ is negative for 0 < x < (2V)^{1/3} and positive for (2V)^{1/3} < x < ∞. Thus S has minimal surface area when x = (2V)^{1/3} and h = (V/4)^{1/3}.

5.6, #5. Let q be the number of chairs ordered. The revenue R as a function of q is

R(q) = \{ \begin{matrix}90q & for 0 ≤ q ≤ 300 \\ (90 − (0.25)(q − 300))q & for 300 < q ≤ 400 \\ 65q & for 400 < q \end{matrix}

Here 90−(0.25)(q −300) is the price of a chair with a reduction of $0.25 for each additional chair sold over 300. The reductions stop when q ≥ 400 with the price stabilizing at $65. So the graph R is the join of a line segment of slope 90, a piece of the parabola 165q − .25q^{2}, and a line segment of slope 65. We find the global extrema of the parabola on [300, 400]. Its critical point is q = 330. Now R(300) = 27000, R(330) = 27225, R(400) = 26000. When 300 ≤ q ≤ 400, the minimal and maximal revenues are $26000 and $27225. When 0 ≤ q ≤ 300, the minimal and maximal revenues are $0 and $27000. When 400 < q, the minimal revenue is $26000; the maximal revenue can be arbitrarily large.

5.6, #6. The cost C of operating the boat is the sum of two costs: the hourly overhead $675 and hourly cost of fuel kv^{3}. Here v is the velocity of the boat in units of miles/hour and k is a positive constant. In fact, k = 0.1 since F (10) = k10^{3} = 100 is given. So C(v) = 675 + 0.1v^{3} in units of dollars/hour. To minimize cost per mile, we minimize C(v)/v = (675/v)+0.1v^{2}, which has units of dollars/mile. (C(v)/v)′ = −(675/v^{2})+0.2v = (0.2v^{3} − 675)/v^{2} has a zero at v^{3} = 3375, that is, at v = 15 miles/hour. This is a global minimum since C(v)/v)′ is negative for 0 < v < 15 and positive for v > 15.