Attempt $1$ $\newline$ Find two positive real numbers such that the sum of the first number squared and the second number is $108$ and their product is a maximum. $\newline$ (Use symbolic notation andractions where needed. Give your answer as a comma separated list of two values.)

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Q. Attempt

1

\newline

Find two positive real numbers such that the sum of the first number squared and the second number is

108

and their product is a maximum.

\newline

(Use symbolic notation andractions where needed. Give your answer as a comma separated list of two values.)

Define Variables: Let's denote the first positive real number as $x$ and the second positive real number as $y$ . According to the problem, we have two conditions: $\newline$ $1$ . The sum of the first number squared and the second number is $108$ , which gives us the equation $x^2 + y = 108$ . $\newline$ $2$ . We want to maximize the product of these two numbers, which is $xy$ . $\newline$ To solve this problem, we can express $y$ in terms of $x$ using the first equation and then find the maximum value of the product $xy$ using calculus or by completing the square. $\newline$ First, let's express $y$ in terms of $x$ using the first equation: $\newline$ $y$ $0$ .
Express $y$ in terms of $x$ : Now we have the product of the two numbers as a function of $x$ : $P(x) = xy = x(108 - x^2).$ To find the maximum value of this product, we need to take the derivative of $P(x)$ with respect to $x$ and set it to zero to find the critical points. Let's calculate the derivative of $P(x)$ : $P'(x) = \frac{d}{dx} [x(108 - x^2)] = 108 - 3x^2.$
Calculate derivative of P(x): Next, we set the derivative equal to zero to find the critical points: $\newline$ $108 - 3x^2 = 0$ . $\newline$ Solving for $x$ gives us: $\newline$ $3x^2 = 108$ , $\newline$ $x^2 = 36$ , $\newline$ $x = \pm6$ . $\newline$ Since we are looking for positive real numbers, we take $x = 6$ .
Find critical points: Now we need to verify that $x = 6$ gives us a maximum value for the product. We can do this by checking the second derivative of $P(x)$ at $x = 6$ or by observing that the function $P(x) = x(108 - x^2)$ is an upside-down parabola, which means the critical point we found is indeed a maximum. $\newline$ The second derivative of $P(x)$ is: $\newline$ $P''(x) = \frac{d^2}{dx^2} [108 - 3x^2] = -6x.$ $\newline$ Evaluating at $x = 6$ gives us: $\newline$ $P''(6) = -6(6) = -36$ , which is less than zero, confirming that $x = 6$ is a maximum.
Verify maximum value: Now that we have the value of $x$ , we can find the corresponding value of $y$ using the equation $y = 108 - x^2$ : $\newline$ $y = 108 - 6^2$ , $\newline$ $y = 108 - 36$ , $\newline$ $y = 72$ .
Find corresponding y value: We have found two positive real numbers, $x = 6$ and $y = 72$ , that satisfy the given conditions. The sum of the first number squared and the second number is $108$ , and their product is a maximum.