find the two numbers whose sum of the squares is a minimum if the product of the numbers is 10.

xy = 10, so y=10/x

z = x^2+y^2 = x^2 + 100/x^2
dz/dx = 2x - 200/x^3
= (2x^2-200)/x^3

dz/dx=0 when x=±10

find two numbers whose product is 16 and whose sum of squares is minimum.

To find the two numbers whose sum of the squares is a minimum if the product of the numbers is 10, we can use the concept of the AM-GM inequality.

Let's assume the two numbers are x and y. We know that the product of the numbers is 10, so we have xy = 10.

To find the sum of the squares, we need to minimize the expression x^2 + y^2.

Using the AM-GM inequality, we have:

(x^2 + y^2)/2 ≥ √(x^2y^2)

Substituting xy = 10, we get:

(x^2 + y^2)/2 ≥ √(10^2)
(x^2 + y^2)/2 ≥ √100
(x^2 + y^2)/2 ≥ 10

To minimize the expression x^2 + y^2, we want the left side of the inequality to be equal to the right side. So,

(x^2 + y^2)/2 = 10

Multiplying both sides by 2, we get:

x^2 + y^2 = 20

Therefore, the two numbers whose sum of the squares is a minimum if the product is 10 are the solutions to the equations:

xy = 10
x^2 + y^2 = 20

To find the two numbers whose sum of the squares is a minimum given that the product of the numbers is 10, we can use a mathematical approach.

Let's denote the two numbers as x and y. We want to minimize the sum of their squares, which can be expressed as x^2 + y^2.

We are given that the product of the numbers is 10, so we can write the equation xy = 10.

To solve this problem, we can use the method of substitution. Rearrange the equation xy = 10 to isolate one of the variables. Let's solve for y:

y = 10 / x

Now, substitute this expression for y in the equation x^2 + y^2:

x^2 + (10 / x)^2

Simplify this equation by expanding the expression:

x^2 + 100 / x^2

To minimize this equation, we need to find the critical points by taking the derivative with respect to x and setting it equal to zero:

d/dx (x^2 + 100 / x^2) = 0

Differentiating the equation gives:

2x - 200 / x^3 = 0

Multiply through by x^3 to clear the fraction:

2x^4 - 200 = 0

Simplify and solve for x:

2x^4 = 200
x^4 = 100
x = ±√(10)

Since we are looking for real numbers, we take the positive square root:

x = √(10)

Now substitute this value back into the equation y = 10 / x to find y:

y = 10 / √(10)
y = √(10)

So the two numbers whose sum of the squares is a minimum, given that their product is 10, are approximately √(10) and √(10).

Please note that these calculations may involve rounding approximations.