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April 1, 2015

April 1, 2015

Posted by **ramesh** on Monday, April 8, 2013 at 7:48am.

- math 12 grade -
**Reiny**, Monday, April 8, 2013 at 8:20amdid you mean x^2 y^5 ???

I will assume that.

x+y = 35 ----> y = 35-x

product = x^2y^5 = x^2(35-x)^5

d(product)/dx = x^2(5)(35-x)^4 (-1) + 2x(35-x)^5

= 0 for a max of product

x^2(5)(35-x)^4 (-1) + 2x(35-x)^5 = 0

x(35-x)^4 [ -5x + 2(35-x) ] = 0

x(35-x)^4 [ -7x + 70 ] = 0

x = 0 or -7x + 70 = 0

so x = 0 or x = 10 , but we wanted positive numbers

so x = 10 and y = 35-10 = 25

The two numbers are 10 and 25

- math 12 grade -
**tyrehon governor**, Friday, April 12, 2013 at 1:43pm55%

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