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November 26, 2014

November 26, 2014

Posted by **Jenny** on Tuesday, December 14, 2010 at 8:41pm.

a) 3481

b) 3600

c) 27,000

d) 32,000

e) 36,000

So Idk if this is right, but I have ....

x+y=60

p=xy^2

p=x(60-x)^2

P'=-2x(60-x)+(60-x^2)

x=.5021 x= 119.498

idk if this is right so far, or if i forgot something. how do I find the max product?

- calculus -
**Reiny**, Tuesday, December 14, 2010 at 9:31pmwhy did you square the more complicated factor, why not square the x

p = x^2(60-x)

= 60x^2 - x^3

dp/dx = 120x - 3x^2

= 0 for a max of p

3x(40 - x) = 0

x = 0 or x = 40

obviously x=0 will produce the minimum product

if x = 40

then 60-40 = 20

so the two numbers are 40 and 20 (with 40 as the number that was squared in the product)

Your way should have worked too, but I notice that you jumped from

.... (60-x)^2 to

.....(60-x^2) , that is an error.

my product is (40^2)(20) = 32 000

btw, your two numbers don't even add up to 60

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