Calculating Dimensions of a Cylindrical Container with Given Volume

[RohanTalkad]

The question reads, "Find the dimensions of a cylindrical tennis ball container which has the volume of V(x)=8πx³+17πx²+10πx+π such that the volume is exactly 825π cm³. Hint: V = πr²h."

To start off, I set V(x)=825π and moved it to the right side, giving

0 = 8πx³+17πx²+10πx-824π.

Factoring pi, we get 0 = π(x-4)(8x²+49x+206),

Since we can't factor the second bracket, here's where I get confused. My inference is that the radius is 4 cm, and the height is muzzled in that unfactorable bracket. However, having the equation for volume (V = πr²h), I get h = 825/16π.

Can someone verify this for me, please?

 

[Mark44]

The question reads, "Find the dimensions of a cylindrical tennis ball container which has the volume of V(x)=8πx³+17πx²+10πx+π such that the volume is exactly 825 cm³. Hint: V = πr²h."

To start off, I set V(x)=825π and moved it to the right side, giving

0 = 8πx³+17πx²+10πx-824π.

Factoring pi, we get 0 = π(x-4)(8x²+49x+206),

Since we can't factor the second bracket, here's where I get confused. My inference is that the radius is 4 cm, and the height is muzzled in that unfactorable bracket. However, having the equation for volume (V = πr²h), I get h = 825/16π.

Looks fine to me, but you should write that number as 825/(16π). Many people would interpret what you wrote as $\frac{825}{16}\pi$.

Can someone verify this for me, please?

 

[andrewkirk]

To start off, I set V(x)=825π

That doesn't sound right. We are given that $V=825$, not $V=825\pi$.

 

[RohanTalkad]

That doesn't sound right. We are given that $V=825$, not $V=825\pi$.

Sorry, I forgot to add 825pi as the volume.