# Product of 2 positive numbers is 16. Find the 2 numbers such that a) their sum is a mminimum b) the sum of one and the square of the other is minimum.

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An easier way:

List the factors of 16

1, 2, 4, 8, 16.

a) minimum sum: 1+16=17, 2+8=10, 4+4=8. The smallest sum made by the appropriately paired factors is 8, so your answer should be 4 and 4.

b) This one we can solve with a little intuition. Which of the factors would be the smallest squared? Answer is 1. The other factor then has to be 16. `16+1^2=17` So the answer is 1 and 16.

If both conditions have to be met, then use mariloucortez's solution, but if a, and b are subquestions, calling for different answers, you've got my answer.

Let x be the first number

y be the second number

The product of the numbers is 16.

The mathematical expression for that statement is:

xy = 16

For the first question, let's minimize their sum. So the equation would be:

S=sum

x+y = S -----> equation to be minimized

From xy = 16, get x in terms of y. So x = `16/y`

Plug-in 16/y in place of x in the equation to be minimized.

16/y + y = S

Make the left side a single fraction.

`(16 + y^2)/y = S`

To minimize the sum get the derivative of the expression first.

Use the quotient rule. d(u/v) = `(vdu - udv)/v^2`

u = `16 + y^2`

v = y

d(`(16+y^2)/y` ) `= ( y(2y) - (16+y^2))/y^2`

`(dS)/dy = (2y^2 - 16 -y^2)/y^2`

`(dS)/dy = (y^2-16)/y^2`

Let the left side be equal to zero.

`0 = (y^2 - 16)/y^2`

Solve for y. Cross multiply.

`0 = y^2 -16`

Factor to solve for y. To factor difference of two square, use:

`a^2 - b^2 = (a-b)(a+b)`

So,

`0 = (y - 4)(y+4)`

y = 4 and y = -4.

But the problem says, "two positive numbers". So, therefore, just consider y=+4.

To solve for x, plug-in 4 in place of y in `x = 16/y`

so x =16/4,

x = 4.

Therefore the numbers are 4 and 4.

To check this get all the factors of 16. Add by pairs and you'll find out the 4 + 4 will give the least sum.

For the "sum of one and square of the other is minimum", get the original equation:

`xy = 16`

Then establish the working equation to be minimized.

`x+y^2 = S`

Do the same process.

`16/y + y^2 = S`

`(16+y^3)/y = S`

`(y(3y^2) -(16+y^3))/y^2 = (dS)/dy`

`(3y^3 -16 -y^3)/y^2 = (dS)/dy`

` (2y^3 - 16)/y^2 = (dS)/dy`

` (2y^3 - 16)/y^2 = 0`

`2y^3 -16 =0`

Factor out 2.

`2(y^3 - 8) = 0`

Equate each factor to 0.

2 = 0 is not true, so disregard.

`y^3 - 8 = 0`

Move -8 to other side and take the cube root.

`y^3 = 8`

`root(3)(y)=root(3)(8)`

y = 2 ---> plug-in this value in place of y in x = 16/y.

x =16/2

x = 8.

So the numbers are 8 and 2.

Let no. be x ,then another number be 16/x , also x >0 .By given condition

`y=16/x+x` , we wish to minimize y.

Differentiate y with respect to x

`dy/dx=16(-1)/x^2+1`

`` For minimum, set dy/dx=0

`-16/x^2+1=0`

`x^2=16`

`x==sqrt(16)`

`x=4`

check `(d^2y)/(dx^2)>0 ` at x=4 for minimum

`(d^2y)/(dx^2)=16(-1)(-2)/x^3`

`(d^2y/dx^2)=32/4^3 >0`

`Thus no.are 4,4`

For second part,please repost problem.