+0  
 
0
235
2
avatar

2. A container in the form of a cylinder, covered on the bottom and the top, is made to hold 8π ft3. Find the dimensions of the cylinder that will require the least amount of material, that is the surface area, to make.

difficulty advanced
Guest Jun 5, 2015

Best Answer 

 #2
avatar+91045 
+10

 A container in the form of a cylinder, covered on the bottom and the top, is made to hold 8π ft3. Find the dimensions of the cylinder that will require the least amount of material, that is the surface area, to make.

 

$$\\V=\pi r^2h=8\pi\\\\
r^2h=8\\\\
r^2=\frac{8}{h}\\\\
r=\sqrt{\frac{8}{h}}\\\\
$Let surface area=A$\\\\
A=2\pi rh+2\pi r^2\\\\
A=2\pi h\sqrt{\frac{8}{h}}+2\pi \frac{8}{h}\\\\
$This will be a max or a min when $\frac{dA}{dh}=0\\\\
A=2\pi h^{0.5}\sqrt{8}+2\pi *8h^{-1}\\\\$$

 

$$\\A=2\pi h^{0.5}\sqrt{8}+2\pi *8h^{-1}\\\\
A=4\sqrt2\pi h^{0.5}+16\pi h^{-1}\\\\
\frac{dA}{dh}=2\sqrt2\pi h^{-0.5}-16\pi h^{-2}\\\\
\frac{d^2A}{dh^2}=-\sqrt2\pi h^{-1.5}+32\pi h^{-3}\\\\$$

 

$$\\$Find stat points $\frac{dA}{dh}=0\\\\
2\sqrt2\pi h^{-0.5}-16\pi h^{-2}=0\\\\
\sqrt2 h^{-0.5}-8 h^{-2}=0\\\\
\sqrt2 h^{1.5}-8=0 \qquad $I mult both sides by $ h^2\\\\
h^{1.5}=\frac{8}{\sqrt2}\\\\
h=[\frac{8}{\sqrt2}]^{2/3}\\\\
h=\frac{2^2}{2^{1/3}}\\\\
h=2^{5/3}\\\\$$

 

 

$$\frac{d^2A}{dh^2}=-\sqrt2\pi h^{-1.5}+32\pi h^{-3}\\$$

$${\mathtt{\,-\,}}{\sqrt{{\mathtt{2}}}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{\left({{\mathtt{2}}}^{\left({\frac{{\mathtt{5}}}{{\mathtt{3}}}}\right)}\right)}^{\left(-{\mathtt{1.5}}\right)}{\mathtt{\,\small\textbf+\,}}{\mathtt{32}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{\left({{\mathtt{2}}}^{\left({\frac{{\mathtt{5}}}{{\mathtt{3}}}}\right)}\right)}^{\left(-{\mathtt{3}}\right)} = {\mathtt{2.356\: \!194\: \!490\: \!192\: \!344\: \!8}}$$

 

GOOD    (I  was hoping  this answer would be positive, otherwise there would be an error)

 

$$If\; h=2^{5/3}\quad $A minimum amount of material will be used$$$

 

$$\\r=\sqrt{\frac{8}{h}}\\\\
r=\sqrt{\frac{2^3}{2^{5/3}}}\\\\
r=\sqrt{2^{4/3}}\\\\
r=2^{4/6}\\\\
r=2^{2/3}\\\\
r=\sqrt[3]{4}\\\\$$

 

So the minimum amount of material will be needed if      $$r=\sqrt[3]{4}\quad and \quad h=\sqrt[3]{32}$$

(The units are in feet)

 

This answer is identical to Alan's but our methods are a bit different     

Melody  Jun 5, 2015
Sort: 

2+0 Answers

 #1
avatar+26329 
+10

 minimum area

 

Notice that the height is twice the radius (or height is the same as the diameter).

.

Alan  Jun 5, 2015
 #2
avatar+91045 
+10
Best Answer

 A container in the form of a cylinder, covered on the bottom and the top, is made to hold 8π ft3. Find the dimensions of the cylinder that will require the least amount of material, that is the surface area, to make.

 

$$\\V=\pi r^2h=8\pi\\\\
r^2h=8\\\\
r^2=\frac{8}{h}\\\\
r=\sqrt{\frac{8}{h}}\\\\
$Let surface area=A$\\\\
A=2\pi rh+2\pi r^2\\\\
A=2\pi h\sqrt{\frac{8}{h}}+2\pi \frac{8}{h}\\\\
$This will be a max or a min when $\frac{dA}{dh}=0\\\\
A=2\pi h^{0.5}\sqrt{8}+2\pi *8h^{-1}\\\\$$

 

$$\\A=2\pi h^{0.5}\sqrt{8}+2\pi *8h^{-1}\\\\
A=4\sqrt2\pi h^{0.5}+16\pi h^{-1}\\\\
\frac{dA}{dh}=2\sqrt2\pi h^{-0.5}-16\pi h^{-2}\\\\
\frac{d^2A}{dh^2}=-\sqrt2\pi h^{-1.5}+32\pi h^{-3}\\\\$$

 

$$\\$Find stat points $\frac{dA}{dh}=0\\\\
2\sqrt2\pi h^{-0.5}-16\pi h^{-2}=0\\\\
\sqrt2 h^{-0.5}-8 h^{-2}=0\\\\
\sqrt2 h^{1.5}-8=0 \qquad $I mult both sides by $ h^2\\\\
h^{1.5}=\frac{8}{\sqrt2}\\\\
h=[\frac{8}{\sqrt2}]^{2/3}\\\\
h=\frac{2^2}{2^{1/3}}\\\\
h=2^{5/3}\\\\$$

 

 

$$\frac{d^2A}{dh^2}=-\sqrt2\pi h^{-1.5}+32\pi h^{-3}\\$$

$${\mathtt{\,-\,}}{\sqrt{{\mathtt{2}}}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{\left({{\mathtt{2}}}^{\left({\frac{{\mathtt{5}}}{{\mathtt{3}}}}\right)}\right)}^{\left(-{\mathtt{1.5}}\right)}{\mathtt{\,\small\textbf+\,}}{\mathtt{32}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{\left({{\mathtt{2}}}^{\left({\frac{{\mathtt{5}}}{{\mathtt{3}}}}\right)}\right)}^{\left(-{\mathtt{3}}\right)} = {\mathtt{2.356\: \!194\: \!490\: \!192\: \!344\: \!8}}$$

 

GOOD    (I  was hoping  this answer would be positive, otherwise there would be an error)

 

$$If\; h=2^{5/3}\quad $A minimum amount of material will be used$$$

 

$$\\r=\sqrt{\frac{8}{h}}\\\\
r=\sqrt{\frac{2^3}{2^{5/3}}}\\\\
r=\sqrt{2^{4/3}}\\\\
r=2^{4/6}\\\\
r=2^{2/3}\\\\
r=\sqrt[3]{4}\\\\$$

 

So the minimum amount of material will be needed if      $$r=\sqrt[3]{4}\quad and \quad h=\sqrt[3]{32}$$

(The units are in feet)

 

This answer is identical to Alan's but our methods are a bit different     

Melody  Jun 5, 2015

6 Online Users

avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details