find the width and the length

Question

The area of a rectangle is 33 ft2

, and the length of the rectangle is 5 ft

 less than twice the width. Find the dimensions of the rectangle.

Answer 1

Hi! This is not a differential equation question since it does not involve equations and derivatives.

It seems more like a precalculus question.

You were given that the area A = 33 ft2 and the length L and width W of a rectangle satisfy the relationship L=2W-5:

A = L*W = 33

L = 2W-5

We replace L into the first equation and solve for W:

(2W-5)*W=33

2 W²-5W-33=0

We use the quadratic equation and obtain W=22/4 and W=-3;

Since W must be positive (it is the length of a rectangle), the only acceptable solution is W=22/4 = 11/2.

We plug replace W into the equation for L, L = 2W-5 = 2(11/2) - 5 = 11-5 = 6.

The dimensions of the rectangle are W = 11/2 and L = 6.

Done.

As an additional step, we may want to check that indeed, A=L*W=6*11/2=33.

I hope this helps! Please let me know if you have any questions! Good luck.