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Some help here would be appreciated.

Julius  Mar 21, 2018
 #1
avatar+87571 
+3

Let x  be th first number and y be the second

So  x + y  =  18   ⇒   y  =  18 - x

 

Call the function that we wish to maximize, M....so we have

 

M  =  xy^2

M = x (18 - x)^2

M =  x (x^2 - 36x + 324)

M = x^3 - 36x^2 + 324x

 

Take the derivative and set to 0

 

M'  = 3x^2 - 72x + 324  =  0     

 

x^2 - 24x + 108  =  0        factor

 

(x - 18) (x - 6)  =  0

 

Set each factor to  0  and solve for x  and we have that x  =  18  or x = 6

 

The second derivative will gives us a min and max for the function

 

Taking the second derivative. we have

 

6x - 72

 

Subbing 18  into this gives a positive....so.... this is a  minimum for the function

Subbing 6  into this gives a negative so this is a  max  for the function

 

So the max product  is  when x  = 6  and y  = 12  =    6 (18 -6)^2   =  6 * 12^2  =  864

 

 

cool cool cool

CPhill  Mar 21, 2018

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