When the product of two positive numbers is 88 what is the maximum sum of their squares

1 Answer | Add Yours

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The product of two numbers is 88. Let one of the numbers be X, the other number is `88/X` . The sum of the square of the two numbers is given by the expression, `S = X^2 + (88/X)^2`

To determine the value of X that maximizes the sum, determine the derivative of S with respect to X.

`(dS)/(dX) = 2X - 15488/X^3`

Solve `(dS)/(dX) = 0`

=> `2X - 15488/X^3 = 0`

=> `X^4 = 7744`

=> `X = 2*sqrt 22`

`S'' = (2X^4+46464)/X^4`

At `X = 2*sqrt 22` , S'' is positive. This indicates that the value of S at `X = 2*sqrt 22` is minimum. The maximum value of the sum of the squares is infinity.

The two numbers with product 88 are X and `88/X` , as X takes on a smaller value, the value of `88/X` increases with a limiting case of 0 and infinity for X and `88/X` respectively. This makes the maximum value of the sum of the square of the two numbers equal to infinity.

We’ve answered 318,989 questions. We can answer yours, too.

Ask a question