You are designing a rectangular poster to contain 50 in^2 of printing with a 4 in. margin at the top and bottom and a 2 inch margin at each side. What overall dimensions will minimize the amount of paper used?

Question

myname · Answer

First. let x, and y be the length and width of the rectangular paper
Then , Area of the print  = 50 = (y-8)(x-4)
x-4 = 50/(y-8)
x= (50/(y-8))+4

Area of the paper = x * y
A(y) = ((50/(y-8))+4)* y --> area of the paper in terms of y
Now, you can use the first derivative test to find out the y that will give the minimum area

$$A \prime(y) = (50/(y-8) + 4) + y(-50/(y-8)^2) $$------>Using the product rule
Next, you need find the critical points by letting A'(y) = 0 and  solving for y
$$0 = (4(y-8)^2-400)/(y-8)^2$$$$0 = (y-8)^2 - 100$$$$0 = y^2 - 16y + 64 -100$$$$0 = y^2 - 16y - 36 $$$$y= -2 $$ or $$y = 18$$
Here, y> 8 from the function A(y). So, -2 cannot be in the domain.
Now , you can use the second derivative test to find whether the area is max or min with the critical point 18.
$$A \prime \prime(y)= 800/(x-8)^3$$$$A \prime \prime(18) = 80$$A''(y) >0,  Therefore y = 18 and x = (50 / (18-8) +4) = 9 will minimize the amount of paper used.