SOLUTION: I am stuck on this problem. Please help. The perimeter of a rectangle is 34 feet and its area is 60 square feet. Find the length and width of the rectangle. Here is what I di

Question 87085This question is from textbook College Algebra Math 107
: I am stuck on this problem. Please help. The perimeter of a rectangle is 34 feet and its area is 60 square feet. Find the length and width of the rectangle.
Here is what I did:
I used the P = 2l + 2w formula
34 = 2l + 2w.
At this point I'm stuck because I'm not sure how to incorporate the area (60 sq. ft.) into the problem. This question is from textbook College Algebra Math 107

Answer by tutorcecilia(2152) (Show Source): You can put this solution on YOUR website!
Use the formula for the Area:
Area = (length)(width)
Area = 60 [plug-in the values that are given]
Sovling for either L or W will be too confusing, so...
.
Use the formula for Perimeter:
P=2(length + width)
P=34 [plug-in the values that are given]
34=2(length + width)
34/2=2(length + width)/2 [solve for either the Width or the Length]
17=length + width
17-width=length+width-width
17-width=length
17-W=L
.
Area=(L)(W)
60=(17-W)(W) [plug-in the values and simplify]
60=17W-w^2 [set the equation equal to zero by adding/subtraction from both sides
60-17W+w^2=-17W+17W-W^2+W^2
60-17W+W^2=0 [re-arrange the terms]
w^2-17W+60=0 [factor]
(W-12)(W-5)=0 [set each factor equal to zero and solve for the W-term]
W-12=0
W=12
or
W-5=0
W=5
.
Since the width is the smallest of the 5 and 12, let 5=width and let 12 = length.
Check by plugging everything back into both equations.
A=LW
60=(12)(5)
60=60 [checks out]
and
P=2(length + width)
34=2(12+5)
34=2(17)
34=34 [also checks out]
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