Page 133 - How To Solve Word Problems In Calculus

We may use the second derivative test to confirm that x = 30
corresponds to a relative minimum (optional).
3600
−3
P (x) = 3600x =

x 3
It is clear that P (30) > 0, so we have a relative minimum at

x = 30. Since it is the only relative extremum on the interval
(0, ∞), the function has its absolute minimum there.
12.

y
x

2x
The amount of material needed to construct the boxis the sum of
the areas of its five sides (it’s an open box). Since the base of the
2
boxis a rectangle x by 2x, its area is 2x . The front and back each
have areas of 2xy and each side, left and right, has an area of xy.
The total area is then
2
A = 2x + 2xy + 2xy + xy + xy
2
= 2x + 6xy
3
Since the volume (l × w × h) is 972 cm ,
(2x)(x)(y) = 972

2
2x y = 972
486
y =
x 2
Substituting into the equation representing A,

486
2
A(x) = 2x + 6x 0 < x < ∞
x 2
2916
2
= 2x +
x
2
= 2x + 2916x −1
Next we differentiate and solve for the critical value:
120