This document provides solutions to three optimization problems involving rectangular advertisements or windows. The first problem minimizes the total area of an advertisement given a fixed printed area, finding the dimensions are a 4.6 inch square printed region within a 6.6 by 6.6 inch advertisement. The second problem maximizes the printed area of an advertisement given a fixed total area of 120 square inches, determining the optimal dimensions are a 4.5 by 6.5 inch advertisement. The third problem maximizes the area of a window with a semicircular top and rectangular body given a fixed amount of trim, concluding the optimal dimensions are a 2 by 4 inch rectangle beneath a semicircle.

1. Solutions to Worksheet for Section 4.5
Optimization Problems
V63.0121, Calculus I
Summer 2010
1. An advertisement consists of a rectangular printed region plus 1 in margins on the sides and
2 in margins on the top and bottom. If the area of the printed region is to be 92 in2 , find the
dimensions of the printed region and overall advertisement that minimize the total area.
Solution. The objective is to minimize the total area while the printed area is fixed.
2 cm
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x
Let the dimensions of the printed region be x and y; then the printed area is P = xy ≡ 92
while the total area is A = (x + 2)(y + 4). From the P equation we can isolate y = 92/x and
substitute it into the A equation to get
92 FOIL 184 184
A(x) = (x + 2) +4 = 92 + 4x + + 8 = 100 + 4x +
x x x
1

2. The domain of A is (0, ∞).
We want to find the absolute minimum value of A. Taking derivatives,
dA 184
=4− 2
dx x
dA √
The equation = 0 gives a single critical point when x = 46 (the other possible one at
√ dx
− 46 is outside the domain). Taking another derivative,
d2 A 368
= 3
dx2 x
√
which is positive along the entire domain of A. So the single local minimum at x =√ 46 is in
fact the global minimum. Therefore, the dimensions of the printed region are x = 46 in by
92 √ √ √
y = √ = 2 46 in. The dimensions of the paper are 46 + 2 in × 2 46 + 4 in.
46
2. An advertisement consists of a rectangular printed region plus 1 in margins on the sides and
1.5 in margins on the top and bottom. If the total area of the advertisement is to be 120 in2 ,
what dimensions should the advertisement be to maximize the area of the printed region?
Solution. Using the notation of the previous problem, the objective is to maximize P while
keeping A fixed.
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120
In this problem, P = xy, but A = (x + 2)(y + 3). Isolating y in A ≡ 120 gives y = −3
x+2
which yields
120 120x
P =x −3 = − 3x
x+2 x+2
2

3. The domain of P is again (0, ∞).
We want to find the absolute maximum value of P . Taking derivatives,
dP (x + 2)(120) − (120x)(1) 240 − 3(x + 2)2
= 2
−3=
dx (x + 2) (x + 2)2
There is a single critical point when
√
(x + 2)2 = 80 =⇒ x = 4 5 − 2
(again, the negative critical point doesn’t count). The second derivative is
d2 P −480
2
=
dx (x + 2)3
√
which is negative all along the domain of P . Hence the unique critical point x = 4 5 − 2 cm
√
is the absolute maximum of P . This means the paper width is 4 5 cm, and the paper length
120 √
is √ = 6 5 cm.
4 5
Here’s another way to do it: Let X and Y be the width and height of the advertisement.
Then we need to maximize P = (X − 2)(Y − 3) subject to the constraint that A = XY ≡ 120.
Substituting Y = 120/X into P gives
120 240
P = (X − 2) −3 = 126 − − 3X
X X
and we need to find the maximum value of P over (0, ∞).
3. A Norman window has the outline of a semicircle on top of a rectangle. Suppose there is
8 + 2π feet of wood trim available. Find the dimensions of the rectangle and semicircle that
will maximize the area of the window.
Solution. Let h and w be the height and width of the window. We have
π
L = 2h + w + w
2
π w 2
A = wh +
2 2
3

4. If L is fixed to be 8 + 2π, we have
16 + 4π − 2w − πw
h= ,
4
so
w π 1 π
A= (16 + 4π − 2w − πw) + w2 = (π + 4)w − + w2 .
4 8 2 8
π
So A = (π + 4)w − 1 + , which is zero when
4
π+4
w= = 4 ft.
1+ π
2
The dimensions are 4ft by 2ft.
Here’s another way to do it: We differentiate the L equation implicitly, keeping in mind it is
equal to a fixed constant:
dh π dh 1 + π/2 π+2
0=2 + 1+ =⇒ =− =− .
dw 2 dw 2 4
Now differentiate A treating h as a function of w:
dA dh π π+2 π w
=h+w + w =h−w + w =h−
dw dw 4 4 4 2
dA
So = 0 when w = 2h, and this is true regardless of what L is fixed to be.
dw
In our case, the relation w = 2h allows to solve L = 8 + 2π for h:
8 + 2π = 2h + (1 + π/2) (2h) = (4 + π)h
So h = 2 and w = 4.
4