Math Problem

[Compnewbie01]

Okay, I took a test today and I just have to know what the answer is to this problem. Think of the graph y = 16-x^2. Now the problem was what dimensions would maximize the perimeter of a rectangle within the curve (assume no negative y values for the rectangle coordinates). I wish there was a way for me to give a better visual of the problem. However, I guess you could say its a rectangle INSIDE the downward parabola where the base of the rectangle is the x-axis.

 

[Goosemaster]

what class? do you know how to integrate?

 

[Compnewbie01]

It is a calculus class and I don't know if integrating would work. Isn't integrating finding the area under the curve? I needed to maximize the perimeter by finding an appropriate equation with the correct variables substituted in (perimeter in terms of either x or y) so that I could find the derivative and use that graph to find the maximum (which would give an x or y value that would maximize the perimeter). Somehow I came up with x = 1 and y = 15 making a perimeter of 32.

 

[her209]

Not in this case bud.

 

[mdchesne]

D+0 > Yi/R + O(w)/n > (herm)Werk

 

[hypn0tik]

What did you get for the perimeter as a function of x?

 

[esun]

Okay, if you have a point of the x-axis as one of the corners of the rectangle, and assume it is symmetric about the y-axis, you have the perimeter is (2 * x) * 2 + 2 * (16 - x^2), since 2 * x is the base of the rectangle and 16 - x^2 is the height of the rectangle. Hence, you take the derivative and get 4 - 4x, which you set equal to zero to find x = 1 (which you can verify is a maximum by looking at the second derivative, -4, which means it is concave up for all x). Hence, the rectangle with maximum perimeter has corners (1, 0), (-1, 0), (1, 15), (-1, 15), since 16 - 1^2 = 15. The perimeter is then 4 + 30 = 34 (not 32). Looks like you got the answer right.