# maximize

• Nov 11th 2008, 08:10 PM
cwarzecha
maximize
1. A hemisphere of radius 7 sits on a horizontal plane. A cylinder stands with its axis vertical, the center of its base at the center of the sphere, and its top circular rim touching the hemisphere. Find the radius and height of the cylinder of maximum volume.

2. A box has a bottom with one edge 2 times as long as the other. If the box has no top and the volume is fixed at V, what dimensions minimize the surface area?
• Nov 11th 2008, 08:42 PM
Soroban
Hello, cwarzecha!

Quote:

A right circular cylinder is inscribed in a hemisphere of radius 7.
Find the radius and height of the cylinder of maximum volume.

This is a side view of the hemisphere and cylinder.

Code:

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Draw line segment $OA.$

The hemisphere has center $O$ and radius $OA = 7.$

The cylinder has the cross-section $ABCD$ with radius $r = OB$ and height $h = AB.$

In right triangle $ABO\!:\;\;h^2+r^2 \:=\:7^2\quad\Rightarrow\quad h \:=\:\sqrt{49-r^2}$ .[1]

The volume of the cylinder is: . $V \;=\;\pi r^2h$ .[2]

Substitute [1] into [2]: . $V \;=\;\pi r^2\left(49-r^2\right)^{\frac{1}{2}}$

And that is the function you must maximize . . .

• Nov 11th 2008, 09:04 PM
cwarzecha
I got the radius to be 6, not sure if that's right, but if it is how would I get the height?
• Nov 14th 2008, 02:54 PM
Soroban
Hello, cwarzecha!

Your work is somewhat off . . .

Quote:

We have:

$h \:=\:\sqrt{49-r^2}$ .[1]

The volume of the cylinder is: . $V \;=\;\pi r^2h$ .[2]

Substitute [1] into [2]: . $V \;=\;\pi r^2\left(49-r^2\right)^{\frac{1}{2}}$

Differentiate and equate to zero . . .

$V' \;=\;\pi\bigg[r\cdot\tfrac{1}{2}(49-x^2)^{-\frac{1}{2}}(-2r) + 2r(49-r^2)^{\frac{1}{2}}\bigg] \:=\:0$

Divide by $\pi\!:\;\;\frac{-r^3}{\sqrt{49-r^2}} + 2r\sqrt{49-r^2} \:=\:0$

Multiply by $\sqrt{49-r^2}\!:\;\;-r^3 + 2r(49-r^2) \:=\:0 \quad\Rightarrow\quad -r^3 + 98r - 2r^3 \:=\:0$

. . $3r^2 - 98r \:=\:0 \quad\Rightarrow\quad r(3r^2 - 98) \:=\:0$

And we have: . $r = 0$, which we reject.

And also: . $3r^2 \:=\:98 \quad\Rightarrow\quad r^2 \:=\:\frac{98}{3} \quad\Rightarrow\quad r \:=\:\sqrt{\frac{98}{3}} \quad\Rightarrow\quad\boxed{ r\:=\; \frac{7\sqrt{6}}{3}}$

Substitute into [1]: . $h \;=\;\sqrt{49 - \left(\frac{7\sqrt{6}}{3}\right)^2} \;=\; \sqrt{\frac{49}{3}} \quad\Rightarrow\quad\boxed{ h \:=\:\frac{7\sqrt{3}}{3}}$