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