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:
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)`
`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`
y = 2 ---> plug-in this value in place of y in x = 16/y.
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
`` For minimum, set dy/dx=0
check `(d^2y)/(dx^2)>0 ` at x=4 for minimum
`Thus no.are 4,4`
For second part,please repost problem.