Learning Activity
Score: KEY
MthSc 106
Section 4.5, Activity 1
1. Show that among all rectangles with an 8 meter perimeter, the one with the largest area is a
square.
a. Draw a picture, label it, and define ALL variables.
y
x
Let x = width of rectangle (m)
y = length of rectangle (m)
A = area of rectangle (m2)
b. Write an equation relating the variables. Then use the given information to write the
equation in terms of only one variable. Give the domain.
Given: 2x + 2y = 8 => 2y = 8 – 2x => y = 4 – x
A = xy
Maximize A( x)  x(4  x)  4 x  x 2
x  0, y  0  4  x  0   x  4  x  4 The domain is 0  x  4 .
c. Find the absolute extreme (and/or where it occurs) and verify the absolute extreme.
A( x)  4  2 x  2(2  x)
A( x)  0 when x  2
x
0
2
4
A(x)
0
4
0
 abs. max.
OR
A
–
+
|
0
|
2
|
4
So, A(x) has an absolute maximum at
x = 2 on [0, 4].
OR
A( x)  2  0 for all x (including x = 2) So, A(x) has an absolute maximum at x = 2.
d. State your conclusion in sentence form.
When x = 2, y = 4 – 2 = 2.
The largest rectangle (in area) with an 8 m perimeter is a square of side 2m.
Learning Activity
Section 4.5, Activity 1
2. A 216 m2 rectangular pea patch is to be enclosed by a fence and divided into two equal parts
by another fence parallel to one of the sides. What dimensions for the outer rectangle will
require the smallest total length of fence? How much fence will be needed?
y
x
y
x
y
x
Let x = length of the divider (m)
y = width of one side (m)
f = amount of fence (m)
y
Given: x(2y) = 216  y = 108/x.
The amount of fence is f = 3x + 4y.
Minimize f (x) = 3x + 432/x, x > 0.
f ( x)  3 
432 3x 2  432 3( x 2  144) 3( x  12)( x  12)



x2
x2
x2
x2
f ( x)  0 when x  12, x  12
f ( x) d.n.e. when x  0
f ( x) 
864
 0 for all x in the domain of f .
x3
f has an absolute minimum at x = 12. When x = 12, y = 9.
The dimensions that require the smallest total length of fence are 18 m by 12 m. The amount
of fence needed is 3(12) + 4(9) = 72 m.