Question Statement

Find two non-negative numbers whose sum is $60$ and whose product is a maximum.

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Background and Explanation

This problem applies optimization techniques from differential calculus to find maximum values subject to constraints. You will need to express the product as a function of a single variable using the given sum constraint, then apply the first and second derivative tests to locate and verify the maximum.

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Solution

Let the first number be $x$ and the second number be $y$.

According to the question, the sum of the two numbers is $60$: \begin{align*} x + y &= 60 \\ y &= 60 - x \end{align*}

Let $P$ represent the product of the two numbers. Substituting equation (1) to express $P$ as a function of $x$ alone: $\begin{array}{rl}P& =xy\\ & =x(60-x)\\ P& =60x-x^2\end{array}$

To find the maximum product, we differentiate $P$ with respect to $x$: $\frac{dP}{dx}=60-2x$

Differentiating again with respect to $x$ to obtain the second derivative: $\frac{d^{2}P}{d{x}^2}=-2$

For critical values:

Set the first derivative equal to zero to find stationary points: $\Rightarrow \begin{array}{rl}60-2x& =0\\ 60& =2x\\ x& =30\end{array}$

Verification of maximum:

At $x=30$, evaluate the second derivative: $\frac{d^{2}P}{d{x}^2}=-2<0$

Since $\frac{d^{2}P}{d{x}^2}<0$, the product $P$ is maximum at $x=30$.

Substituting $x=30$ back into equation (1) to find the second number:

Therefore, the two non-negative numbers are both $30$.

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Key Formulas or Methods Used

• Constraint equation: $x+y=60$ (used to eliminate one variable)
• Objective function: $P=xy=60x-x^2$ (product expressed in terms of $x$)
• First derivative test: $\frac{dP}{dx}=0$ for finding critical points
• Second derivative test: $\frac{d^{2}P}{d{x}^2}<0$ confirms a local maximum

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Summary of Steps

1. Define variables: Let the numbers be $x$ and $y$, with constraint $x+y=60$.
2. Formulate objective function: Express product as $P(x)=x(60-x)=60x-x^2$.
3. Find critical points: Compute $\frac{dP}{dx}=60-2x$, set to zero, and solve to get $x=30$.
4. Verify maximum: Check that $\frac{d^{2}P}{d{x}^2}=-2<0$ to confirm a maximum exists at $x=30$.
5. Calculate second number: Substitute $x=30$ into $y=60-x$ to find $y=30$.