A rectangle has a perimeter of 80 cm. If its width is x, express its length and its area in terms of x. What is the maximum area of the rectangle?
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Originally Posted by ep78 A rectangle has a perimeter of 80 cm. If its width is x, express its length and its area in terms of x. What is the maximum area of the rectangle? Let x be the width and y be the height. Thus 2x+2y=80. Solving for y, we get x+y=40 and y=40-x. Let A(x) be an area of function. A(x) = x(40-x). That's just the length times the width all in terms of x. Now differentiate and find the critical point.
Originally Posted by Jameson Let x be the width and y be the height. Thus 2x+2y=80. Solving for y, we get x+y=40 and y=40-x. Let A(x) be an area of function. A(x) = x(40-x). That's just the length times the width all in terms of x. Now differentiate and find the critical point. so would i plug in any number into x(40-x) or is there a certain one?
Ooops! This is pre-calc. I was throwing in some Calculus terms. You could graph this area function and visually spot the maximum. I will say that a square is the best shape at maximizing area when the perimeter is constrained.
oh haha, okay thank you very much!
Originally Posted by ep78 so would i plug in any number into x(40-x) or is there a certain one? Differentiate then find the critical numbers. (Where the derivative = 0) dy/dx=40-2x 40-2x=0 x=20
Originally Posted by Len Differentiate then find the critical numbers. (Where the derivative = 0) dy/dx=40-2x 40-2x=0 x=20 Look at title. I made the same mistake. (Pre-calc)
Originally Posted by Jameson Look at title. I made the same mistake. (Pre-calc) Ahh sorry my internet crashed when I was replying and when it came back online I pressed refresh and it sent.
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