calculus help? anyone?

Question

a right triangle whose hypotenuse is the square root of three meters long is revolved about one of its legs to generate a right circular cone. Find the radius, height, and volume of the cone of GREATEST volume that can be made this way.

thanks Wal C!

Answer 1

Let one leg be x m long
Then the other leg is √(3 - x²)

Two possiblilites

Case 1 Rotate about leg with length x m

So r = √(3 - x²) m and h = x m

V = ⅓πr²h
= ⅓π(√(3 - x²))² x
= π/3(3x - x³)

dV/dx = π(1 - x²)
= 0 for stationary points

x = 1 ( ignoring -1 as x is a positive quantity)

When x = 1- dV/dx > 0 and when x = 1+ dV/dx < 0 So there is a local maximum at x = 1

Thus maximum volume when
height = 1 m
radius = √(3 - 1²) = √2 m

And then maximum volume is 2π/3 m³

Case 2 Rotate about leg with length √(3 - x²) m

So r = x m and h = √(3 - x²) m

V = ⅓πr²h
= ⅓πx² √(3 - x²)
= ⅓π√(3x^4 - x^6)

dV/dx = ⅓π (12x³ - 6x^5) * ½ * 1/√(3x^4 - x^6)
= πx³(2 - x²)/[x² √(3 - x²)]
=πx(2 - x²)/[x² √(3 - x²)]
= 0 for stationary points
So x = 0, √2 (ignoring -√2 as r cannot be negative)

Again when x = √2, V is maximum (V = 0 when x = 0) so the solution is again as above

ie radius is √2 m, height is 1 m and volume is 2π/3 m³

Answer 2

very easy!