Question

We want to build a greenhouse that has a half cylinder roof of radius r and height r mounted horizontally on top of four rectangular walls of height h as shown in the figure. We have 200π m² of plastic sheet to be used in the construction of this structure. Find the value of r for the greenhouse with the largest possible volume we can build.

Answer 1

$V=2r^2+\frac{\pi r^2h}{2};$ $S_{surface}=2\pi r^2+2r^2+6rh;$ We applied differentiation like this:

1) $\frac{dV}{dh}+\frac{dV}{dr}=0:$

$\frac{\pi r^2}{2}\frac{dV}{dh}+4r+ \pi rh=0;$

$\frac{dV}{dh}=-\frac{2(4+\pi h)}{\pi r};$

2) $\frac{dS}{dh}+\frac{dS}{dr}=0:$

$6r\frac{dS}{dh}+4\pi r+4r+6h=0;$

$6r(-\frac{2(4+\pi h)}{\pi r})+4\pi r+4r+6h=0;$

$-12(4+\pi h)+4\pi^2 r+4\pi r+6\pi h=0;$

$-48-12\pi h+4\pi^2 r+4\pi r+6\pi h=0;$

$-48-6\pi h+4\pi^2 r+4\pi r=0;$

$h=\frac{2\pi^2 r+2\pi r-24}{3\pi};$

There is given $S=\pi rh=200$ :

$\pi r\times\frac{2\pi^2 r+2\pi r-24}{3\pi}=200;$

$r\times(\pi^2 r+\pi r-12)=300;$

$r\times(r\times(1+\frac{1}{\pi})-\frac{12}{\pi^2})=300;$

$r_1=\frac{12+\sqrt{144+1200\pi ^4+1200\pi ^3}}{2\pi ^2+2\pi }$ $r_2=\frac{12-\sqrt{144+1200\pi ^4+1200\pi ^3}}{2\pi ^2+2\pi }$

Here is $r_2$ is negative. So r would get the value of $r_1$ . And can find V:

$V=2r^2+\frac{\pi r^2h}{2};$

$V=2r^2+\frac{\pi r^2 \times\frac{2\pi^2 r+2\pi r-24}{3\pi}}{2};$

$V=2r^2+\frac{ r^2\times(2\pi^2 r+2\pi r-24)}{6};$

$V=r^3(\frac{ \pi^2 +\pi }{3})-2r^2=0;$

$V=r^2\times(r\times(\frac{ \pi^2 +\pi }{3})-2)=0;$ where, $r=\frac{12+\sqrt{144+1200\pi ^4+1200\pi ^3}}{2\pi ^2+2\pi };$