35 - 37 Solved problems in maxima and minima

Print Area:
$(x - 2)(y - 3) = 24$

$y = \dfrac{24}{x - 2} + 3$
 

Page area:
$A = xy$

$A = x \left( \dfrac{24}{x - 2} + 3 \right)$

$A = \dfrac{24x}{x - 2} + 3x$

$\dfrac{dA}{dx} = \dfrac{(x - 2)24 - 24x(1)}{(x - 2)^2} + 3 = 0$

$\dfrac{-48}{(x - 2)^2} + 3 = 0$

$-48 + 3(x - 2)^2 = 0$

$x = \sqrt{48/3} + 2$

$x = 6 \, \text{ in}$
 

$y = \dfrac{24}{6 - 2} + 3$

$y = 9 \, \text{ in}$
 

Dimensions: 6 in × 9 in           answer

Given perimeter:
$P = b + 2(h - r) + \pi r$

$P = b + 2h - 2r + \pi r$

Where:
$b = 2r$

$r = \frac{1}{2}b$

 

Thus,
$P = b + 2h - b + \frac{1}{2}\pi b$

$P = 2h + \frac{1}{2}\pi b$

$\dfrac{dP}{db} = 2 \dfrac{dh}{db} + \frac{1}{2}\pi = 0$

$\dfrac{dh}{db} = -\frac{1}{4}\pi$
 

Light is most if area is maximum:
$A = \frac{1}{2}\pi r^2 + b(h - r)$

$A = \frac{1}{2}\pi (\frac{1}{2}b)^2 + b(h - \frac{1}{2}b)$

$A = \frac{1}{8}\pi b^2 + bh - \frac{1}{2}b^2$

$A = \frac{1}{8}(\pi - 4)b^2 + bh$

$\dfrac{dA}{db} = \frac{2}{8}(\pi - 4)b + b \dfrac{dh}{db} + h = 0$

$\frac{1}{4}\pi b - b - \frac{1}{4}\pi b + h = 0$

$h = b$
 

∴   breadth = height           answer

From Solution of Problem 36
$\dfrac{dh}{db} = -\frac{1}{4}\pi$

$r = \frac{1}{2} b$
 

Half amount of light is equivalent to half of the area.
$A = \frac{1}{4}\pi r^2 + b(h - r)$

$A = \frac{1}{4}\pi (\frac{1}{2}b)^2 + b(h - \frac{1}{2}b)$

$A = \frac{1}{16}\pi b^2 + bh - \frac{1}{2}b^2$

$A = \frac{1}{16}(\pi - 8)b^2 + bh$

$\dfrac{dA}{db} = \frac{2}{16}(\pi - 8)b + b \dfrac{dh}{db} + h = 0$

$\frac{1}{8}\pi b - b - \frac{1}{4}\pi b + h = 0$

$-(1 + \frac{1}{8}\pi)b + h = 0$

$h = (1 + \frac{1}{8}\pi)b$

$height = (1 + \frac{1}{8}\pi) \, \times \, breadth$           answer