# The product of two numbers is 1296. What is the maximum sum of the numbers ?

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Let the numbers be x and y.

Given that the product of the of the numbers is 1296.

x*y = 1296

==> y= 1296/ x

Let the sum of the numbers be f(x) = x+ y.

We will wrtie f(x) as a function of x.

==> f(x) = x + ( 1296/x)

==> f(x) = (x^2 + 1296)/x

Now we need to find the extreme value for f(x).

Let us find the first derivative.

==> f'(x) = ( 2x*x - x^2+ 1296))/(x^2)

= ( 2x^2 - x^2 + 1296) / x^2

= ( x^2 - 1296)/x^2

==> f'(x) = (x^2 - 1296)/x^2

==> (x^2 - 1296 = 0

==> x^2 = 1296

==> x= 36

Then the function has extreme value at x= 36.

==> y= 36.

**Then, the maximum sum of the two numbers is 36+ 36 = 72.**

Let the numbers be x and y. Now we are given that the product of the numbers is 1296.

So x*y = 1296

=> x = 1296 / y

Now the sum of the numbers is x + y

=> 1296/ y + y

We have to maximize 1296/ y + y

Let's find the derivative of 1296/ y + y

=> -1296 / y^2 + 1

Equate this to 0

=> -1296 / y^2 + 1 = 0

=> - 1296/ y^2 = -1

=> y^2 = 1296

=> y = +36 and -36

For y = 36, 1296/ y + y = 72. This is the largest value.

**Therefore the largest value that the sum of the two numbers can be is 72.**

Since the product of two numbers is 1296, we assume one number x and the other is 1296/x.

Let the sum s(x) of the numbers be x+1296/x.

So s(x) = x+1296/x.

By common sense it could be seen that s(x) has no maximum as s(x) = x+(1296/x) approach infinity as x-->0. Alsoone of the number 1296/x --> infinity as x --> 0.

Simmilarly it has no minimum if you consider negative numbers also, as x+1296/x approach -ve infinity as x--> infinity.

However if it is the minimum sum s(x) that is required of product of two inegers, then we get s(x) by differentiating and equation to zero. s'(x) = 0 gives 1-1296/x^2 = 0 giving x^2 =1296. Or x= sqrt1296 = 36. So s(x) = 36+1296/36 = 72 is the minimum positive sum of the numbers whose product is 1296. As s"(x) = 2*1296/x^3 and s"(36) = 1296/36^3 >0.

Again there is -36*-36 = 1296 and -36+-36 = -72

Also 1296 = -1296*-1. So (-1296)+(-1) = -1297 is the minimum sum of integer product.

Let's note the numbers as x and y:

We'll write their product:

x*y = 1296

y = 1296/x

We'll write the function of sum:

S(x) = x + 1296/x

The sum is maximum when the first derivative of the function is cancelling.

S'(x) = 0

S'(x) = (x + 1296/x)'

S'(x) = 1 - 1296/x^2

We'll put S'(x) = 0

1 - 1296/x^2 = 0

We'll multiply by x^2:

x^2 - 1296 = 0

We'll re-write the difference of squares:

x^2 - 36^2 = 0

(x - 36)(x + 36) = 0

x - 36 = 0

x = 36

x = -36

y = 1296/36

y = 36

If the sum of the numbers is maximum, we'll reject the negative value for x.

x+ y = 72 (1)

We'll isolate y to the left side and

**The sum is maximum for x = 36 and y = 36**

**S = 36 + 36**

**S = 72**