Question

How to do no1 and 2 ?

Answer 1

1.

`f(x) = x(x-4) = x^2 - 4x`

`f'(x) = 2x - 4`

`f'(0) = 2(0) - 4 = -4`

Slope of perpendicular to tangent, -1/(-4) = 1/4.

We want the line tangent to f(x) with this slope:

`1/4 = f'(x) = 2x - 4`

`17/4 = 2x`

`x = 17/8`

`y = x(x-4) = (17)/(8) \cdot (-(15)/(8)) = -(255)/(64)`

Answer: B(17/8, -255/64)

Looks like you got it right.

2.

The cylinder volume is
`2\pi r h`, here

`81 \pi= \pi x^2 h`

`h = (81)/(x^2)`

(a) Total surface area is circular base plus cylinder side plus the lateral cone surface The lateral cone surface area, surface length l, radius r is
`\pi r l`.

`A = \pi x^2 + 2\pi x\cdot(81)/(x^2) + \pi x(2x)`

`A = 3 \pi x^2 + (162 \pi)/(x) = 3 \pi(x^2+ 54/x)`

(b)

`A' =0 = 3\pi(2x-54/x^2)`

`54 = 2x^3`

`x^3 = 27`

`x=3`

`A = 3 \pi(x^2+ 54/x) = 3 \pi(9 + 18) = 81 \pi`

Answer: 81π