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Math 11: A Rectangle Word Problem using the Quadratic Formula

Formula for perimeter of a rectangle: P = 2w + 2l, where w=width and l=length

Formula for area of a rectangle: A = wl

30 = 2w+2l

40 =wl

In the area formula, isolate for l:

l = 40/w

Substitute this into the perimeter formula:

30 = 2w + 2(40/w)
30 = 2w + 80/w
30w = 2w^2 + 80
0 = 2w^2 -30w + 80

This doesn’t factor nicely, so just use the quadratic formula to find w. If you need help with the quadratic formula let me know.

You could also just use something like Wolfram Alpha to graph it and find the solutions:
Graph of w roots, showing the 2 possible values for w.

Anyway, you should find that w=(1/2)(15-sqrt(65)), which is the same thing as 3.47 in decimal form.

Alternately, you could have used the other answer from the quadratic formula, w=(1/2)(15+sqrt(65)), which is the same thing as 11.53 in decimal form.

So w=3.47 and l=11.53