A rectangle has dimensions of 3x and 5 - 2 X

Question

what is the maximum area of the rectangle?

what value of x gives the maximum area?

 

Initial equation is 3x(5-2x) 

Then -6x^2+15x

Do I use quadratic formula? 

Answer 1

Mary anne,

 

A(x) = 3x(5-2x) = 15x - 6x².

 

If you have had Calculus, set the first derivative to zero, dA(x)/dx = 15 - 12x = 0, so x = 5/4 = 1.25.

And A91.25) = 9.375

 

If you only know Algebra, then first notice that x>0 and (5-2x)>0 must hold to have any positive length sides to the rectangle. So your solution must satisfy  0 < x < 2.5,

 

Now plot the area function A(x) in this range [0,2.5] and see where it maximizes graphically, then check it.  Or numerically,

    A(0)    = 0

    A(0.5)  = 6

    A(1)     = 9

    A(2)     = 6

 

So you can see that the function peaks to the left or right of x = +1.  So you can try  0.8, 0.9, 1.1, 1.2, 1.3 to narrow it down to between 1.2 & 1.3.  So now try 1.21, 1.22 ... 1.29. Then you can go to the thousands place to get x = 1.25.

 

Or you can reduce the work by using the average value of x values just to the left and right of the optimum point. So if you found that the neighborhood is [0,2.5] initially, then the half-way point is x=1.25 to try next. Coincidentally, it hits it right away, but you don't know that yet. So try the midpoints of  [1.2, 1.25] and [1.25, 1.3] = 1.225 & 1.275, etc, and see that their A values are below x=1.25, and keep repeating to get the accuracy you need in the solution for x.