Find two numbers whose difference is 16, and the sum of whose squares a minimum.

Thanks....I figured it out to be (8, -8) also. Thanks again.

The numbers 🤣 are 3 and 5

Well, finding two numbers whose difference is 16 is not a difficult task. Let's call the first number "x" and the second number "y". So, we have the equation x - y = 16.

Now, you're asking for the sum of their squares to be a minimum. To minimize the sum of the squares, we need to find the two numbers that are closest to each other. The numbers that are closest to each other are the ones that have the smallest difference.

In this case, since x - y = 16, the two numbers that are closest to each other are the ones whose difference is 16. So, if we take x = 8 and y = -8, their difference is indeed 16.

Now, let's calculate their sum of squares:

x^2 + y^2 = 8^2 + (-8)^2 = 64 + 64 = 128.

Therefore, the two numbers that have a difference of 16 and whose sum of squares is a minimum are 8 and -8, and the minimum sum of their squares is 128.

I hope this puts a smile on your face!

To find two numbers whose difference is 16 and the sum of whose squares is a minimum, we can follow these steps:

Step 1: Let's assume the two numbers as x and y. Since their difference is given as 16, we can write the equation: x - y = 16.

Step 2: Now, let's express one of the numbers in terms of the other. Solving the equation from Step 1 for x, we get: x = y + 16.

Step 3: We need to minimize the sum of their squares, so let's form the equation for that. The sum of their squares can be written as: S = x^2 + y^2.

Step 4: Substitute the value of x from Step 2 into the equation from Step 3: S = (y + 16)^2 + y^2.

Step 5: Simplify the equation in Step 4: S = y^2 + 32y + 256 + y^2.

Step 6: Combine like terms in the equation from Step 5: S = 2y^2 + 32y + 256.

Step 7: To minimize the sum of squares, we need to find the vertex of the quadratic equation. The vertex of a quadratic equation in the form ax^2 + bx + c is given by x = -b/2a.

In our case, the quadratic equation is 2y^2 + 32y + 256, so the vertex is found using the formula: y = -32/(2*2) = -8.

Step 8: Substitute the value of y into the equation from Step 2 to find x: x = y + 16 = -8 + 16 = 8.

So, the two numbers whose difference is 16 and the sum of whose squares is a minimum are 8 and -8.

minimize x^2+y^2 where x-y=16

x=8 y=-8

That is, f(x) = x^2 + (x-16)^2
f' = 4(x-8)
f'=0 when x=8

x = first number

y = second number

x - y = 16

that is equivalent of

y = x - 16

x ^ 2 + y ^ 2 = x ^ 2 + ( x - 16 ) ^ 2 =

x ^ 2 + x ^ 2 - 2 * x * 16 + 16 ^ 2 =

2 x ^ 2 - 32 x + 256

Quadratic equation a x ^ 2 + b x + c

vhen a is positive in point :

x = - b / 2 a

have MINIMUM

x = - b / 2 a = - ( - 32 ) / ( 2 * 2 ) = 32 / 4 = 8

When x = 8 minimum value of 2 x ^ 2 - 32 x + 256 =

2 * 8 ^ 2 - 32 * 8 + 256 = 2 * 64 + 256 + 256 = 128

x = 8

y = x - 16 = 8 - 16 = - 8

x ^ 2 + y ^ 2 = 8 ^ 2 + ( - 8 ) ^ 2 = 64 + 64 = 128