# 38 - 40 Solved problems in maxima and minima

**Problem 38**

A cylindrical glass jar has a plastic top. If the plastic is half as expensive as glass, per unit area, find the most economical proportion of the jar.

**Solution:**

$V = \pi r^2 h$

$0 = \pi r^2 \dfrac{dh}{dr} + 2\pi rh$

$\dfrac{dh}{dr} = -\dfrac{2\pi rh}{\pi r^2}$

$\dfrac{dh}{dr} = -\dfrac{2h}{r}$

m = price per unit area of glass

½ m = price per unit area of plastic

k = total material cost per jar

$k = m (2\pi rh + \pi r^2) + \frac{1}{2}m (\pi r^2)$

$k = 2m \pi rh + \frac{3}{2}m \pi r^2$

$\dfrac{dk}{dr} = 2m\pi \left( r\dfrac{dh}{dr} + h \right) + 3m \pi r = 0$

$2 \left( r\dfrac{dh}{dr} + h \right) + 3r = 0$

$2 \left[ r \left( -\dfrac{2h}{r} \right) + h \right] + 3r = 0$

$2h = 3r$

$h = \frac{3}{2}r$

Height = 3/2 × radius of base *answer*

**Problem 39**

A trapezoidal gutter is to be made from a strip of tin by bending up the edges. If the cross-section has the form shown in Fig. 38, what width across the top gives maximum carrying capacity?

**Solution:**

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

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

$h = \frac{1}{2} \sqrt{4a^2 - (b - a)^2}$

Capacity is maximum if area is maximum:

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

$A = \frac{1}{2}(b + a) \left( \frac{1}{2} \sqrt{4a^2 - (b - a)^2} \right)$

$A = \frac{1}{4}(b + a) \sqrt{4a^2 - (b - a)^2}$ (*take note that 'a' is constant*)

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

$\sqrt{4a^2 - (b - a)^2} = \dfrac{b^2 - a^2}{\sqrt{4a^2 - (b - a)^2}}$

$4a^2 - (b - a)^2 = b^2 - a^2$

$4a^2 - b^2 + 2ab - a^2 = b^2 - a^2$

$2b^2 - 2ab - 4a^2 = 0$

$b^2 - ab - 2a^2 = 0$

$(b + a) (b - 2a) = 0$

For b + a = 0; b = -a (meaningless)

For b - 2a = 0; b = 2a (*ok*)

Use b = 2a *answer*