Page 112 - How To Solve Word Problems In Calculus

Step4
Differentiating this expression is a little messy due to the
presence of the radical. There is a simple trick that can be
used to make the calculation easier. Instead of minimizing d,
2
2
we minimize d . Since the value of d is smallest when d is
smallest, minimizing d 2 will lead to the same point that
minimizes d.
2
For convenience, let D = d .
2
2
D(x) = x + (6 − 3x) − 4x − 6(6 − 3x) + 13
2
2
= x + 36 − 36x + 9x − 4x − 36 + 18x + 13
2
= 10x − 22x + 13
D (x) = 20x − 22

0 = 20x − 22

11 60 33 27
x = y = 6 − 3x = − =
10 10 10 10

A glance at the figure should convince you that a mini-
11
mum distance certainly exists. Since x = is the only critical
10

11 27
number, the point , is the point on the line closest
10 10
to (2, 3).

EXAMPLE 14
2
A rectangular poster, which is to contain 50 in of print, must
have margins of 2 in on each side and 4 in on the top and bot-
tom. What dimensions will minimize the amount of material
used?

Solution
Step1
Although we could let x and y represent the dimensions
of the poster, it turns out that labeling the inner rectangle
containing the print leads to a simpler solution.

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