Two numbers have a sum of 72.  What is their product if it is a maximum?

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giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

Let's note the numbers as x and y:

x+ y = 72 (1)

We'll isolate y to the left side and we'll get:

y = 72 - x

We'll write their product:

x*y = x(72 - x)

We'll note the expression of the product:

f(x) = x(72 - x)

We'll remove the brackets:

f(x) = -x^2 + 72x

Now, the value of the product is maximum (because the coefficient of x^2 is negative) if the derivative of f(x) is cancelling (equals zero).

We'll calculate f'(x):

f'(x) = -2x + 72

We'll put f'(x) as 0:

-2x + 72 = 0

We'll divide by -2:

x - 36 = 0

x = 36

The other number is y = 72 - x.

y = 72 - 36

y = 36

The numbers x and y are equal.

f(x) = x*y

f(36) = 36*36

f(36) = 1296

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neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

The sum of the two numbers is 72.

 So we assume x is a number and 72-x is the other number.

Their product  P(x) = x(72-x).

We have to find the number x such that x(72-x) is maximum.

We know by calculus that P(x) is maximum  for x= c,  if P'(c) = 0 or {x(72-x}' = 0 and p"(c) < 0.

P'(c) = 0 gives : {x(72-x)}' = 0. Or {72x-x^2}' = 0, or 72-2x= 0, or 72=2x.  So c = x = 72/2 = 36.

P"(x) = (72-2x)' = -2.  So P"(c) = -2 which is < 0.

Therefore, if x= c = 36, then substituting into the function P(x):

 x(72-x) --> 36(72-36) = 36^2 is maximum.

So the maximum product numbers whose sum is 72  is 36^2 = 1296 .

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