Find two positive numbers whose sum is 110 and whose product is maximized.

1. Problem Description

Find two positive numbers whose sum is 110 and whose product is maximized.

2. Solution Steps

Let the two positive numbers be $x$ and $y$. We are given that $x + y = 110$, and we want to maximize the product $P = xy$.

From the equation $x + y = 110$, we can express $y$ in terms of $x$: $y = 110 - x$.

Substitute this expression for $y$ into the product $P$:

$P = x(110 - x) = 110x - x^2$.

To maximize $P$, we can find the vertex of the parabola represented by the quadratic equation $P(x) = -x^2 + 110x$. The x-coordinate of the vertex is given by $x_v = -\frac{b}{2a}$, where $a = -1$ and $b = 110$.

$x_v = -\frac{110}{2(-1)} = \frac{110}{2} = 55$.

Now, we can find the value of $y$ by substituting $x = 55$ into the equation $y = 110 - x$:

$y = 110 - 55 = 55$.

Thus, the two numbers are $x = 55$ and $y = 55$.

To check that this gives a maximum, we can take the second derivative of $P(x) = -x^2 + 110x$.

$P'(x) = -2x + 110$

$P''(x) = -2$

Since $P''(x) < 0$, the function $P(x)$ is concave down, and we have found a maximum.

3. Final Answer

The two positive numbers are 55 and 55.