Question

A pair of secret numbers is hidden in the linear equation: 14x + 2024. The sum of these two positive numbers equals the rate at which this expression changes with respect to x. Among all such pairs, find the ones that keep their sum of squares as low as possible. What are these numbers?

Answer 1

7 and 7

Answer 2

The problem asks us to find two positive numbers, say a and b, based on conditions derived from the linear equation 14x + 2024.

First, let's determine the sum of these two numbers. The problem states that "The sum of these two positive numbers equals the rate at which this expression changes with respect to x." The given expression is E = 14x + 2024. The rate at which this expression changes with respect to x is its derivative with respect to x: $\frac{dE}{dx} = \frac{d}{dx}(14x + 2024) = 14$ So, the sum of the two secret numbers a and b is 14: $a + b = 14 \quad \text{(Equation 1)}$

Next, we need to find the pair of numbers that keeps their sum of squares as low as possible. We want to minimize the sum of squares, S = a^2 + b^2.

From Equation 1, we can express b in terms of a: $b = 14 - a$ Substitute this into the expression for S: $S = a^2 + (14 - a)^2$ To find the minimum value of S, we differentiate S with respect to a and set the derivative to zero: $\frac{dS}{da} = \frac{d}{da}[a^2 + (14 - a)^2]$ $\frac{dS}{da} = 2a + 2(14 - a)(-1)$ $\frac{dS}{da} = 2a - 2(14 - a)$ $\frac{dS}{da} = 2a - 28 + 2a$ $\frac{dS}{da} = 4a - 28$ Set the derivative to zero to find critical points: $4a - 28 = 0$ $4a = 28$ $a = 7$ To confirm that this value of a corresponds to a minimum, we find the second derivative of S with respect to a: $\frac{d^2S}{da^2} = \frac{d}{da}(4a - 28) = 4$ Since $\frac{d^2S}{da^2} = 4 > 0$, the function S has a minimum at a = 7.

Now, substitute a = 7 back into Equation 1 to find b: $7 + b = 14$ $b = 14 - 7$ $b = 7$ Both numbers are positive (7 and 7), which satisfies the problem's condition.

Thus, the two secret numbers are 7 and 7.