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.