among all right circular cones with a slant height of 33, what are the dimensions (radius and height) that…

Explanation:

Step1: Recall volume and slant - height formulas

The volume of a right - circular cone is $V=\frac{1}{3}\pi r^{2}h$, and the slant height $l$ is related to the radius $r$ and height $h$ by $l^{2}=r^{2}+h^{2}$. Given $l = 33$, so $r^{2}=33^{2}-h^{2}=1089 - h^{2}$.

Step2: Substitute $r^{2}$ into the volume formula

Substitute $r^{2}=1089 - h^{2}$ into $V=\frac{1}{3}\pi r^{2}h$, we get $V(h)=\frac{1}{3}\pi(1089 - h^{2})h=\frac{1}{3}\pi(1089h - h^{3})$.

Step3: Find the derivative of $V(h)$

Using the power rule, $V^\prime(h)=\frac{1}{3}\pi(1089 - 3h^{2})$.

Step4: Set the derivative equal to zero

Set $V^\prime(h)=0$, then $\frac{1}{3}\pi(1089 - 3h^{2}) = 0$. Divide both sides by $\frac{1}{3}\pi$ to get $1089 - 3h^{2}=0$. Rearrange to $3h^{2}=1089$, then $h^{2}=363$, and $h = 11\sqrt{3}$.

Step5: Find the radius

Since $r^{2}=1089 - h^{2}$, substitute $h = 11\sqrt{3}$ into it. $r^{2}=1089-363 = 726$, so $r=\sqrt{726}=11\sqrt{6}$.

Answer:

The height must be $11\sqrt{3}$ and the radius of the base must be $11\sqrt{6}$.