Answer:
The dimensions to maximize the area is 50 by 100
Step-by-step explanation:
Let x= the length of the side perpendicular to the barn
and 200-2x= length of the side parallel to the barn
Area = x(200-2x) (1)
Area =-[tex]x^{2}[/tex]+200x
The formula for the w-value of maximum is: -b/2a
thus our value will be:
200/4 =50
and substitue x = 50 into (1), we have: 200-2(20) =100
Thus the dimensions to maximize the area is 50 by 100
