Page 157 - Calculus Workbook For Dummies

141
Chapter 8: Using Differentiation to Solve Your Problems

25
That’s about 15 ounces. The can will be 2 1 $ .63 or about 3 ⁄4 inches wide and - . 1 63 or
1
π $ . 1 63
about 3 ⁄4 inches tall. Isn’t that nice? The largest can has the same width and height and would
1
thus fit perfectly into a cube. Geometric optimization problems frequently have results where
the dimensions have some nice, simple mathematical relation to each other.
b . . . What dimensions will minimize the cost of the frame? The dimensions are 4’3’’ wide and
5’1’’ high. The minimum cost is $373.
1. Draw a diagram with variables (see the following figure).

y

2x

2. a. Express the thing you want to minimize, the cost.
Cost = _ length of curved frame $ _ cost per linear foot +
i
i
i
_ length of straight frame $ _ cost per linear footi
x 25 + _
= ^ π h ^ h 2 x + y 2 ^ i 20h
= 25 π x + 40 x + 40 y
b. Relate the two variables to each other.
Area = Semicircle + Rectangle
π x 2
20 = + 2 xy
2
c. Solve for y and substitute.
π x 2 cost 25 π x + 40 x + 40 y
=
2 xy 20 -
=
2 10 π x
^ h
20 π x 2 C x = 25 π x + 40 x + 40 c x - m
y = - 4
2 x 4 x
10 π x = 25 π x + 40 x + 400 - 10 π x
= - x
x 4 400
= 15 π x + 40 x + x
3. Find the domain.
x > 0 is obvious. And when x gets large enough, the entire window of 20 square feet in area
will be one big semicircle, so
π x 2
20 =
2
40 = π x 2
40
2
x = π
40
x = π
. . 3 57
Thus, x must be less than or equal to 3.57.