MATH 180 FALL 1995

ASSIGNMENT #35

5.6, #8. The inscribed rectangle has corners at (9, 0), (x, 0), (x, \sqrt{x}), and (9, \sqrt{x}). Here 0 ≤ x ≤ 9. In particular, the rectangle has base length 9 − x and height \sqrt{x}. So the area A = (9 − x)\sqrt{x} and the perimeter P = 2(9 − x) + 2\sqrt{x}. Now

A′ = \frac{9}{2\sqrt{x}} − \frac{3\sqrt{x}}{2} = \frac{3(3 − x)}{2\sqrt{x}}, P′ = −2 + \frac{1}{\sqrt{x}}.

A has a critical point at x = 3, P has a critical point at x = 1/4, and both points are in [0, 9]. Since A(0) = 0, A(3) = 6\sqrt{3} = 10.39, A(9) = 0 and P (0) = 18, P (1/4) = 18.5, P (9) = 6, A and P are maximal respectively at x = 3 and x = 1/4.

5.6, #10. The distance to be minimized is \sqrt{(x − a)^{2} + y^{2}}, where a = 2 or \sqrt{8} and where y^{2} = 1−(x/3)^{2}. Minimizing distance is the same as minimizing the square of the distance. So we minimize

f(x) = (x − a)^{2} + y^{2} = (x − a)^{2} + 1 − (x/3)^{2}

on the interval [−3, 3]. Now f′(x) = 2(x − a) − 2(x/3))(1/3) = (2/9)(7x − 8a). Thus f has a critical point at x = 9a/8. When a = 2, the critical point x = 9/4 is in [−3, 3]. Since f(−3) = 27, f(9/4) = 1.625, f(3) = 3, x = 9/4 is the global minimum. When a = \sqrt{8}, the critical point x = 9/\sqrt{8} = 3.182 is not in [−3, 3]. Since f(−3) = 35.97, f(3) = 2.03, x = 3 is the global maximum.

5.6, #12. The length of the pipe is given by Pythogoras’s theorem: L = \sqrt{x^{2} + 1} + \sqrt{(4 − x)^{2} + 16}. This is to be minimized for 0 ≤ x ≤ 4.

\begin{aligned}L′ & = \frac{2x}{2\sqrt{x^{2} + 1}} + \frac{2(4 − x)(−1)}{2\sqrt{(4 − x)^{2} + 16}} \\ & = \frac{x\sqrt{(4 − x)^{2} + 16} − (4 − x)\sqrt{x^{2} + 1}}{\sqrt{x^{2} + 1}\sqrt{(4 − x)^{2} + 16}}. \end{aligned}

L′ = 0 when x\sqrt{(4 − x)^{2} + 16} = (4 − x)\sqrt{x^{2} + 1}. Squaring both sides gives

x^{2}((4 − x)^{2} + 16) = (4 − x)^{2}(x^{2} + 1).

After simplification this becomes 16x^{2} = (4 − x)^{2}. Since x and 4 − x are positive, taking square roots gives 4x = 4 − x and x = 4/5. This is the global minimum since L(0) = 6.66, L(4/5) = 6.40, L(4) = 8.12.