To solve the problem of finding the smallest value of the sum of the squares of two positive numbers whose total is 16, let’s denote the two numbers as x and y.
Given the constraint:
x + y = 16
We want to minimize the sum of their squares:
S = x² + y²
We can use the constraint to rewrite the sum of squares in terms of a single variable. From the equation x + y = 16, we can express y as:
y = 16 – x
Now, substitute this expression for y into the sum of squares:
S = x² + (16 – x)²
Now expand the equation:
S = x² + (256 – 32x + x²)
S = 2x² – 32x + 256
This is a quadratic equation in terms of x. The sum of squares S can be simplified further:
S = 2(x² – 16x + 128)
The vertex of a parabola described by the equation ax² + bx + c can be calculated at x = -b/(2a). Here, a = 2 and b = -32, so:
x = -(-32)/(2 imes 2) = 32/4 = 8
Since both numbers are positive, we also calculate y:
y = 16 – x = 16 – 8 = 8
Thus, the two numbers that yield the minimum sum of squares are both 8. Now plug these values back into the sum of squares:
S = 8² + 8² = 64 + 64 = 128
In conclusion, the smallest value of the sum of the squares of two positive numbers, given that their sum is 16, is:
128