Question

Question #1be57

Answer 1

The cylinder has radius, `r = 11/3` and `h = 1.5`

Explanation:

If you put the center of the cone on the y axis and look at the cross-section of the cone in the x-y plane, it is a triangle that intersects the y axis at 4.5 and the x axis at 5.5; the equation of that line is:

`y = -9/11x + 4.5" [1]"`

We want to find the point where the cylinder intersects this line such that its volume is maximized:

`V = pir^2h`

But we are looking at these solid objects as cross-sections in the x-y plane so `x = r and y = h`:

`V = pix^2y" [2]"`

Substitute the right side of equation [1] for y in equation [2]:

`V = pix^2(-9/11x + 4.5)`

`V = (-9pi)/11x^3 + (9pi)/2x^2`

Compute the first derivative with respect to x:

`(dV)/dx = (-27pi)/11x^2 + (18pi)/2x`

Set the first derivative equal to zero:

`(-27pi)/11x^2 + 9pix = 0`

Divide by `9pix`:

`(-3)/11x + 1 = 0`

(Please notice that we got rid of the root x = 0 but that is clearly a minimum)

Solve for x:

`x = 11/3`

Do the second derivative test:

`(d^2V)/dx^2 = (-74pi)/11x + 9pi|_(x=11/3) = -(47pi)/3`

It is clearly a maximum.

Substitute `x = 11/3` into equation [1]:

`y = -9/11(11/3) + 4.5`

`y = 1.5`

Translating back to r and h:

The cylinder has radius, `r = 11/3` and `h = 1.5`