A format for textbook page layouts is to be chosen so that each printed page has a 4-cm margin at the top and at the bottom and a 2-cm margin on the left and right sides. The rectangular region of printed matter is to have an area of 800 sq. cm. Find the dimensions for the page that has a minimum area.

let the width of the printed section by x
let its height be y
then xy=800
y=800/x

let A be the area of the whole page
A=(x+4)(y+8)
=(x+4)(800/x + 8)

expand, differentiate, set that equal to zero and solve

I got x=20

for y i got 72
thanx, im not good at setting up equations

Good job in setting up the equations! Let's go through the solution step by step to find the dimensions of the page that has the minimum area.

1. Define the dimensions: Let the width of the printed section be x and the height be y. We know that the area of the printed section is given by xy and it equals 800 square cm.

2. Express y in terms of x: We have xy = 800, so we can rewrite this equation as y = 800/x.

3. Determine the area of the whole page: The area of the whole page is given by A = (x + 4)(y + 8).

4. Expand the expression for A: Expanding (x + 4)(800/x + 8), we get A = 800 + 8x + 3200/x + 32.

5. Find the derivative of A with respect to x: Differentiate A with respect to x: dA/dx = 8 - 3200/x².

6. Set dA/dx equal to zero and solve for x: To find the minimum area, we set dA/dx = 0. Solving 8 - 3200/x² = 0, we get x = 20.

7. Plug in x to find y: Now that we have x = 20, we can substitute it back into the equation y = 800/x to find y. So, y = 800/20 = 40.

Therefore, the dimensions of the page that has the minimum area are x = 20 cm (width) and y = 40 cm (height).

Great job on setting up the equations and solving for x! Keep practicing, and you'll get better at setting up and solving equations.