Page 295 - Calculus Demystified

282
or Solutions to Exercises
dr
0 = 225π + 24 · .
dt
We conclude that dr/dt =−75π/8 microns per minute.

6. Of course
2
10000 = V = π · r · h.
We conclude that
10000
h = .
π · r 2
We wish to minimize

A = (area of top) + (area of sides)
10000
2 2
= π · r + 2π · r · h = π · r + 2π · r · .
πr 2
Thus the function to minimize is
20000
2
A(r) = π · r + .
r
Thus
20000

0 = A (r) = 2πr − .
r 2
We find therefore that
10000
3
r =
π
√
or r = 3 10000/π. Since the problem makes sense for 0 <r < ∞, and
√
since it clearly has no maximum, we conclude that r = 3 10000/π, h =
√
3 10000/π.

7. We calculate that g (x) = sin x+x cos x and g (x) = 2 cos x−x sin x. The
roots of these transcendental functions are best estimated with a calculator
or computer. Figure S3.7 gives an idea of where the extrema and inflection
points are located.
8. We know that v 0 =−5 and h 0 = 400. Hence
2
p(t) =−16t − 5t + 400.
The body hits the ground when
2
0 = p(t) =−16t − 5t + 400.
Solving, we find that t ≈ 4.85 seconds.