Can someone please help me with this question???
Determine the dimensions of the cone of maximum volume that can be inscribed in a sphere with radius 3?
Hello, philipsach1!
Did you make a sketch?
[I] Determine the dimensions of the cone of maximum volume
that can be inscribed in a sphere with radius 3?Code:A * * * * /:\ * * / : \ * * / : \ * / :3 \ * / : \ * * / O* \ * * / 3* : *3 \ * / * y: * \ B *- - - - + - - - -* C * : x * * : * * * *
The center of the sphere is $\displaystyle O.$
$\displaystyle \Delta ABC$ is the side view of the cone.
Note that: $\displaystyle OA = OB = OC = 3.$
The volume of a cone is: .$\displaystyle V \;= \;\frac{1}{3}\pi r^2h$ [1]
In the diagram, we see that: $\displaystyle x =\text{ radius, and }y + 3 =\text{ height}$
So we have: .$\displaystyle V \;= \;\frac{1}{3}\pi x^2(y + 3)$ [2]
The diagram also tells us that: .$\displaystyle x^2 + y^2\:=\:3^2\quad\Rightarrow\quad x^2 \,= \,9 - y^2$
Substitute into [2]: .$\displaystyle V \;= \;\frac{1}{3}\pi(9 - y^2)(y + 3)$
Can you finish up?
Hello again, philipsach1!
The rest is "simple" Calculus . . . but okay.
We have: .$\displaystyle V \;= \;\frac{1}{3}\pi(9 - y^2)(y + 3)$
How do we maximize the volume?
We equate its derivative to zero and solve.
The function is: .$\displaystyle V \;= \;\frac{\pi}{3}\left(-y^3 - 3y^2 + 9y + 27\right)$
So: .$\displaystyle V' \;= \;\frac{\pi}{3}\left(-3y^2 - 6y + 9\right) \;=\;-\pi(y^2 + 2y - 3) \;= \;0$
And we have the quadratic: .$\displaystyle y^2 + 2y - 3 \;= \;0$
. . which factors: .$\displaystyle (y - 1)(y + 3) \;= \;0$
. . and has roots: .$\displaystyle y \:= \:1,\;-3$
Since we want positive dimensions: .$\displaystyle y \:=\:1\quad\Rightarrow\quad x = 2\sqrt{2}$
. . $\displaystyle \begin{array}{ccc}\text{The radius is: } r \:= \:x \:= \:2\sqrt{2} \\ \\ \text{The height is: }h \:= \:y + 3 \:= \:4\end{array}
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