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January 30, 2015

January 30, 2015

Posted by **Mary** on Wednesday, March 6, 2013 at 1:10pm.

I came up with this so far:

x + y = 15

xy^2 is the maximum

derivative of xy^2= 2xyy' + y^2

Now how do I solve this ^ after I set it to zero? I am stuck on that. Thank you so much

- Calculus Optimization Problem -
**Dr. Jane**, Wednesday, March 6, 2013 at 1:13pmYou need to substitute

y = 15-x

x(15-x)^2

x(225 -30x+x^2)

225x -30x^2 + x^3

Now you can take the derivative and set it equal to zero.

- Calculus Optimization Problem -
**Mary**, Wednesday, March 6, 2013 at 1:19pmThank you! I solved it out, and I got x=5 and y= 10 with a product of 500. Is this correct

- Calculus Optimization Problem -
**Dr. Jane**, Wednesday, March 6, 2013 at 1:30pmI agree. You are welcome.

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