minimize ammount of paper used

Question

A rectangular page is to contain 8 square inches of print. The margins at the top and bottom of the page are to be 2 inches wide. The margins on each side are to be 1 inch wide .Find the dimensions of the page that will minimize the amount of paper used. (Let x represent the width of the page and let y represent the height

Answer 1

Let A=xy be the area of the page. Then the area of the printed material is 8 and with the margins included gives the following. 8=(x-2)(y-4). This gives 8=xy-2y-4x+8. Now 0=xy-2y-4x and hence 4x=y(x-2). So y=4x/(x-2). Therefore A=4x^2/(x-2). Take the derivative of A with respect to x to get dA/dx=(8x(x-2)-4x^2)/(x-2)^2. To find the minimum set dA/dx to 0. Hence you get 8x^2-16x-4x^2=0. Solve for x to get x=4 and then y=8 as the minimum dimensions for the page.

Answer 2

Hi Ryan,

As you preferred, let x represent the width of the page and let y represent the height.

We want to minimize the amount of the paper used, i.e. the area of the paper .

Area of the paper = A=x*y <-------- area of a rectangle is the product of its width and its height ( * is used to represent multiplication)

Now we need to express the area of print (8 square inches) in terms of x and y.

Print region is also a rectangle with the dimensions:

width of the print region = x- 2*1 = x-2 <--------- width of the paper (x) minus two times the margin (1 inch) on one side

height of the print region = y- 2*2 = y-4 <---------- height of the paper (y) minus two times the margin (2 inches) at the top (or the bottom) of the page

area of the print region = (x-2) * (y-4) <----------- we substitute the width and the height of print region, which we already found in the previous steps, in the area formula of the rectangle (product of the width and height)

We want print area to be equal to 8 square inches. So,

8 = (x-2)*(y-4)

8 = x*y-4x-2y+8 (cancel 8 from both sides of the equation and then write - 4x on the other side as 4x)

4x =x*y-2y

4x = y * (x-2) (y is the common term on right hand side. Write it outside the paranthesis)

Now we can express y in terms of x as y = 4x ⁄ (x-2).

Remember that we want to minimize the area of the paper A= x*y

In this formula substitute 4x/ (x-2) for y.

Then we obtain the area of the paper in terms of x as A=x* (4x/(x-2)) = 4x² / (x-2)

Since we want to minimize A, we are going to take its derivative with respect to x and equate it to zero. i.e. (dA/dx =0)

dA/dx = ( (4x² )´ *(x-2) - (4x² ) * (x-2)´) / (x-2)² <------ derivative of rational function

0 = (8x *(x-2) - (4x² ) * 1) / (x-2)²

0 = (8x² -16x-4x² ) / (x-2)²

0 = (4x² -16x)/ (x-2)²

0 = 4x² -16x <--------numerator is equal to zero ( keep in mind that the values of x must not make the denominator zero, i.e. x≠2)

0 = 4x* (x-4)

So x=0 or x=4. Since x is the width of the paper it can't be zero so x=4.

We know from that y = 4x/(x-2)

when we substitute 4 for x, we find the height of the paper y = 4*4/(4-2) =16/2 =8

So x =4, y=8 minimize the amount of the paper, i.e. area of the paper.

Now let's check our work. Remember that the area of print region must be 8 square inches. According to the values we found for x and y, the width of the print region is x-2=4-2=2 and the height of the print region is y-4=8-4=4 and the print area is (x-2)*(y-4)=2*4=8. The condition for print area is satisfied.

I hope this helped you and wish you a great semester at school. Feel free to ask any questions you may have regarding this solution by clicking the "Add Comment" prompt. Best!