Question

If x and y are two nonnegative numbers and the sum of twice the first ( x ) and three times the second ( y ) is 60, find x so that the product of the first and cube of the second is a maximum.

Answer 1

If we translate the word problems to mathematical equation,
 
                             2x + 3y = 60

The second equation is,
 
                            P = xy³

From the first equation, we get the value of y in terms of x.
  
                             y = (60 - 2x) / 3

Then, substitute the expression of y to the second equation,
 
                            P = x (60-2x) / 3
                            P = (60x - 2x²) / 3 = 20x - 2x²/3

We derive the equation and equate the derivative to zero.

                           dP/dx = 0 = 20 - 4x/3

The value of x from the equation is 15.

Hence, the value of x for the value of the second expression to be maximum is equal to 15. 
                               

Answer 2

The value of x for the value of the second expression to be maximum is equal to 10.                            

We have to translate the word problems to mathematical equation so we get,

$2x + 3y = 60$

The second equation is,

The product of the first and cube of the second is a maximum.

Therefore we get, $P = xy^3$

What is the maximum value?

The maximum value of a function is the place where a function reaches its highest point, or vertex, on a graph.

From the first equation, we get the value of y in terms of x.

$y = (60 - 2x) / 3$

Then, substitute the expression of y to the second equation,

$P = x ((60-2x) / 3)^3$

  $P = ((60x - 2x^2) / 3 )^3=( 20x - 2x^2/3)^3$

We have to find the maximum value therefore we get,

We derive the equation and equate the derivative to zero.

$dP/dx = 0 =( 20 - 2x/3)^3$

The value of x from the equation is 10.

Therefore, the value of x for the value of the second expression to be maximum is equal to 10.

To learn more about the maximum value visit:

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