# 25 - 27 Solved problems in maxima and minima

**Problem 25**

Find the most economical proportions of a quart can.

**Solution:**

$V = \dfrac{1}{4}\pi d^2 h = \, \text{ 1 quart }$

$\dfrac{1}{4}\pi \left[ d^2 \dfrac{dh}{dd} + 2dh \right] = 0$

$\dfrac{dh}{dd} = -\dfrac{2h}{d}$

Total area (closed both ends):

$A_T = 2(\frac{1}{4}\pi d^2)+ \pi d \, h$

$A_T = \frac{1}{2}\pi d^2 + \pi d \, h$

$\dfrac{dA_T}{dd} = \pi d + \pi \left[ d\dfrac{dh}{dd} + h \right] = 0$

$d d\dfrac{dh}{dd} + h = 0$

$d + d\left( -\dfrac{2h}{d} \right) + h = 0$

$d = h$

Diameter = height *answer*

**Problem 26**

Find the most economical proportions for a cylindrical cup.

**Solution:**

$V = \dfrac{1}{4}\pi d^2 h$

$0 = \dfrac{1}{4}\pi \left[ d^2 \dfrac{dh}{dd} + 2dh \right]$

$\dfrac{dh}{dd} = -\dfrac{2h}{d}$

Area (open one end):

$A = \frac{1}{4}\pi d^2 + \pi d \, h$

$\dfrac{dA}{dd} = \frac{1}{2} \pi d + \pi \left[ d\dfrac{dh}{dd} + h \right] = 0$

$\frac{1}{2}d + d\dfrac{dh}{dd} + h = 0$

$\frac{1}{2}d + d\left( -\dfrac{2h}{d} \right) + h = 0$

$\frac{1}{2}d = h$

$r = h$

Radius = height *answer*

**Problem 27**

Find the most economical proportions for a box with an open top and a square base.

**Solution:**

$V = x^2 y$

$0 = x^2 \, y' + 2xy$

$y' = -2y/x$

Area:

$A = x^2 + 4xy$

$dA/dx = 2x + 4(x\,y' + y) = 0$

$2x + 4 \, [ \, x(-2y/x) + y \, ] = 0$

$2x - 8y + 4y = 0$

$2x = 4y$

$x = 2y$

Aide of base = 2 × altitude *answer*

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