**Problem 41**

In Problem 39, if the strip is L in. wide, and the width across the top is T in. (T < L), what base width gives the maximum capacity?

**Solution:**

$h^2 + \left( \dfrac{T - a}{2} \right)^2 = \left( \dfrac{L - a}{2} \right)^2$

$h = \sqrt{\dfrac{(L - a)^2}{4} - \dfrac{(T - a)^2}{4}}$

$h = \frac{1}{2} \sqrt{(L - a)^2 - (T - a)^2}$

$h = \frac{1}{2} \sqrt{L^2 - 2La + a^2 - T^2 + 2Ta - a^2}$

$h = \frac{1}{2} \sqrt{L^2 - T^2 - 2(L - T)a}$

Area:

$A = \frac{1}{2}(a + T)h$

$A = \frac{1}{2}(a + T)\left( \frac{1}{2} \sqrt{L^2 - T^2 - 2(L - T)a} \right)$

$A = \frac{1}{4}(a + T) \sqrt{L^2 - T^2 - 2(L - T)a}$ (*note that L and T are constant*)

$\dfrac{dA}{da} = \dfrac{1}{4} \left[ (a + T) \dfrac{-2(L - T)}{2\sqrt{L^2 - T^2 - 2(L - T)a}} + \sqrt{L^2 - T^2 - 2(L - T)a} \right] = 0$

$\sqrt{L^2 - T^2 - 2(L - T)a} = \dfrac{(a + T)(L - T)}{\sqrt{L^2 - T^2 - 2(L - T)a}}$

$L^2 - T^2 - 2(L - T)a = (a +T)(L - T)$

$L^2 - T^2 - 2La + 2Ta = La - Ta + TL - T^2$

$L^2 - TL = 3La - 3Ta$

$3(L - T)a = L(L - T)$

$a = \frac{1}{3}L$

Base = 1/3 × length of strip *answer*

**Problem 42**

From a strip of tin 14 inches a trapezoidal gutter is to be made by bending up the sides at an angle of 45°. Find the width of the base for greatest carrying capacity.

**Solution:**

$\sin 45^\circ = \dfrac{h}{\dfrac{14 - a}{2}}$

$h = \dfrac{14 - a}{2} \sin 45^\circ$

$h = \dfrac{14 - a}{2} \left( \dfrac{1}{\sqrt{2}} \right)$

$h = \dfrac{14 - a}{2\sqrt{2}}$

Area:

$A = A_1 + 2A_2 $

$A = ah + 2(\frac{1}{2}h^2)$

$A = a \left( \dfrac{14 - a}{2\sqrt{2}} \right) + \left( \dfrac{14 - a}{2\sqrt{2}} \right)^2$

$A = \dfrac{7a}{\sqrt{2}} - \dfrac{a^2}{2\sqrt{2}} + \dfrac{(14 - a)^2}{8}$

$\dfrac{dA}{da} = \dfrac{7}{\sqrt{2}} - \dfrac{a}{\sqrt{2}} - \dfrac{14 - a}{4} = 0$

$\dfrac{7}{\sqrt{2}} - \dfrac{a}{\sqrt{2}} - \dfrac{7}{2} + \dfrac{a}{4} = 0$

$\left( \dfrac{1}{4} - \dfrac{1}{\sqrt{2}} \right) a = \dfrac{7}{2} - \dfrac{7}{\sqrt{2}}$

$a = 3.17 \, \text{ in }$ *answer*