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:

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

$17/4 = 2x$

$x = 17/8$

$y = x(x-4) = \frac{17}{8} \cdot (-\frac{15}{8}) = -\dfrac{255}{64}$

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

Looks like you got it right.

2.

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

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

$h = \dfrac{81}{x^2}$

(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 $\pi r l$.

$A = \pi x^2 + 2\pi x\cdot\dfrac{81}{x^2} + \pi x(2x)$

$A = 3 \pi x^2 + \dfrac{162 \pi}{x} = 3 \pi(x^2+ 54/x)$

(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