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. 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.
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
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.
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.
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