Question 1148245
.
Find two nonnegative numbers whose sum is 9 and so that the product of one
number and the square of the other number is a maximum 10.
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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Formulation of your post is &nbsp;<U>NOT &nbsp;CORRECT</U> &nbsp;and makes a few sense, if any . . . 


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Meanwhile, &nbsp;an interesting problem may stay behind it  --- so I edited it, &nbsp;to make sense and to create really interesting problem.


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;My editing is as follows.


<pre>
              Prove that if the sum of two non-negative real number "x" and "y" is 9,  x + y = 9,

              then the product  {{{x*y^2}}}  has the local maximum value of 108.

              Find the values of x and y that provide this maximum.
</pre>

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;With this reformulation, &nbsp;the solution is below.



<U>Solution</U>


<pre>
If  x+y = 9,  then  y = 9-x,  and the function  f(x,y) = {{{x*y^2}}}  is


    f(x,y) = {{{x*y^2}}} = {{{x*(9-x)^2}}}.


So, we need to find the maximum of the function of x  g(x) = {{{x*(9-x)^2}}}.


To find the maximum of the function g(x), take its derivative and equate it to zero.


The derivative is

    g'(x) = {{{(9-x)^2}}} - 2x*(9-x) = {{{81 - 18x + x^2}}} - {{{(18x - 2x^2)}}} = {{{81 - 36x + 3x^2}}}.


Equating it to zero, you get

    {{{81 - 36x + x^2}}} = 0,  which is equivalent to  {{{x^2 - 12x + 27}}} = 0.


Left side is factorable

    (x-9)*(x-3) = 0.


and two solution of the quadratic equation are x= 9  and  x= 3.


The value x= 9 provides the local minimum of the function f(x), while x= 3 provides the local maximum.


At x= 3,  y= 9-3 = 6,  and the function g(x,y) = {{{3*6^2}}} = 108.



    {{{graph( 330, 330, -4, 20, -20, 120,
          x*(9-x)^2
)}}}


        Plot  g(x) = x*(9-x)^2
</pre>


Solved.