[CALCULUS] A large rectangular picture frame is to have an area of 120 sq. inches, with 2-inch margins at the top and sides and 1-inch margin at the bottom.......

Question

A large rectangular picture frame is to have an area of 120 sq. inches, with 2-inch margins at the top and sides and 1-inch margin at the bottom. What dimensions of the frame will give the largest area for the picture inside?

Answer 1

Let dementions of frame x and y. Then dimentions of picture is (x - 4) and (y - 3) and area A = (x - 4)(y - 3) =

xy - 3x - 4y + 12; A = 132 - 3x - 4y. Now xy = 120; y = 120/x and we get A(x) = 132 - 3x - 480/x;

A'(x) = - 3 + 480/x² = 0; x² = 160; x = ±4√10. We have 2 critical numbers.

Now A''(x) = - 960/x³; A''(4√10) = - 960/(4√10)³ < 0 and by 2nd derivative test we have at x = 4√10 maximum

y = 120/(4√10) = 3√10

Answer: for x = 4√10 inch and y = 3√10 inch

Answer 2

Let x=width of the picture and y=height of the picture.

The area of the frame is (x+2)(y+3)=120

y= [120/(x+2)]-3

xy=[120x/(x+20}-3x is to be maximized

Set the derivative equal to zero: {[(120(x+2)-120x]/(x+2)²}-3=0

This collects to x²+4x-76=0 and the roots are 2±sqrt(80) which gives size of the picture to be close to 11" square and the frame then is larger by 2' in width and 3" in height.