Question

The number 72 is to be represented as the sum of two positive numbers, such that the product of one of the numbers by the cube of the other is a maximum. what are the two numbers?

Answer 1

f you find a problem in finding out the factors of a number, here's one simple method.
Let's apply that for number 72.

Divide the number by 1 => 72/1=72 ( 1 and 72)
Divide the number by 2 => 72/2=36 (2 and 36)
Divide the number by 3 => 72/3=24 (3 and 24)
Divide the number by 4 => 72/4=18 (4 and 18)
Divide the number by 6 => 72/6=12 (6 and 12)
Divide the number of 8 => 72/8=09 (8 and 9)

So we have total 12 factors.

Let's select number 1 by default. Then the possible cases are
1 2 36
1 3 24
1 4 18
1 6 12
1 8 9

Now let's go with number 2 by default. We will not select 1 in any case now. Then the possible cases are
2 3 12
2 4 9
Now let's go with number 3, leaving 1 and 2. The only possible case is 3 4 6

So, we can express the number 72 as the product of 3 natural numbers in 8 ways.

Answer 2

The value of  the two numbers are 18 and 54.

Step-by-step explanation:

Given:

The sum of two positive numbers is 72.

The product of one of the numbers by the cube of the other is a maximum.

To Find:

The value of  the two numbers.

Solution:

Let two numbers are $x\ and\ y$.

As given,the sum of two positive numbers is 72.

$x+y=72$

$y=72-x$  -------- equation no.01.

As given,the product of one of the numbers by the cube of the other is a maximum.

$f(x)=x( 72-x)^{3}$

$f^{'} (x)=1\times(72-x)^{3} +x\times 3(72-x)^{2}(0-1)$

        $=(72-x)^{3} - 3x(72-x)^{2}$

For maximum  value $f'(x) =0$

$0=(72-x)^{3} - 3x(72-x)^{2}$

$(72-x)^{3} -=3x(72-x)^{2}$

$72-x=3x$

$4x=72$

$x=18$

Putting the value of x in equation no.01, we get.

$y=72-18$

  $=54$

Thus,the value of  the two numbers are 18 and 54.

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