Problem 59 - Algebra and Trigonometry

Question

Find the local maximum and minimum values of the function and the value of x at which each occurs. State each answer correct to two decimal places. $$ U(x)=x \sqrt{6-x} $$

Step-by-Step Solution

Answer

Local maximum is 5.66 at x = 4; local minima are 0 at x = 0 and x = 6.

Step 1 - Find the Critical Points

First, we need to find the critical points of the function. The critical points occur where the derivative of the function is zero or undefined. To find the derivative, use the product rule on the function \( U(x) = x \sqrt{6-x} \).Let \( u = x \) and \( v = \sqrt{6-x} \), then \( u' = 1 \) and \( v' = \frac{-1}{2\sqrt{6-x}} \).Using the product rule \((uv)' = u'v + uv'\), we have:\[U'(x) = 1 \times \sqrt{6-x} + x \times \left( \frac{-1}{2\sqrt{6-x}} \right)\]Simplify:\[U'(x) = \sqrt{6-x} - \frac{x}{2\sqrt{6-x}}\]Combine terms over a common denominator:\[U'(x) = \frac{2(6-x) - x}{2\sqrt{6-x}} = \frac{12 - 3x}{2\sqrt{6-x}}\]Set the numerator equal to zero:\[12 - 3x = 0 \3x = 12 \x = 4\]So, the critical point is \( x = 4 \).

Step 2 - Evaluate the Endpoints and Critical Points

The domain of the function \( U(x) = x\sqrt{6-x} \) is \( 0 \leq x \leq 6 \) because \( 6-x \geq 0 \).Evaluate \( U(x) \) at the critical point \( x = 4 \) and the endpoints \( x = 0 \) and \( x = 6 \). For \( x = 0 \):\[ U(0) = 0 \sqrt{6-0} = 0 \]For \( x = 4 \):\[ U(4) = 4\sqrt{6-4} = 4 \times \sqrt{2} = 4 \times 1.41 = 5.64 \approx 5.66 \]For \( x = 6 \):\[ U(6) = 6\sqrt{6-6} = 6 \times 0 = 0 \]

Step 3 - Determine Local Maximum and Minimum Values

Compare the values obtained from evaluating the critical point and endpoints.- At \( x = 0 \), \( U(0) = 0 \)- At \( x = 4 \), \( U(4) = 5.66 \)- At \( x = 6 \), \( U(6) = 0 \)The local maximum value is \( 5.66 \) at \( x = 4 \).Since the endpoints \( x = 0 \) and \( x = 6 \) both yield the minimum value of \( 0 \), we recognize they are local minima.

Key Concepts

Critical Points

In calculus, critical points are the values of \( x \) where the derivative of a function is either zero or undefined. They are crucial because they can indicate where a function’s graph changes its direction, possibly leading to a local maximum or minimum.
To find critical points, follow these steps:

• First, differentiate the given function to obtain the derivative.
• Set the derivative equal to zero and solve for \( x \).
• Check where the derivative is undefined, especially if it involves division or square roots.

For example, in the function \( U(x) = x\sqrt{6-x} \), using the product rule yields its derivative. By setting \( U'(x) = \frac{12 - 3x}{2\sqrt{6-x}} = 0 \), and solving, we find a critical point at \( x = 4 \). It's essential to check endpoints of the domain if applicable since function behavior can change there as well.

Product Rule

The product rule is a derivative rule used when differentiating a function that is the product of two simpler functions. It states that if you have a function \( U(x) = f(x)g(x) \), the derivative is given by \( U'(x) = f'(x)g(x) + f(x)g'(x) \).
Consider the components of the function and apply the rule as follows:

• Take the derivative of the first function, \( f(x) \), and multiply it by the second function, \( g(x) \).
• Add that to the first function, \( f(x) \), multiplied by the derivative of the second function, \( g'(x) \).

For \( U(x) = x\sqrt{6-x} \), set \( u = x \) and \( v = \sqrt{6-x} \). Then, calculate derivatives: \( u' = 1 \) and \( v' = \frac{-1}{2\sqrt{6-x}} \). Applying the product rule gives \( U'(x) = \sqrt{6-x} - \frac{x}{2\sqrt{6-x}} \), which helps us find critical points.

Local Maximum and Minimum

Local maximum and minimum are points where a function reaches its highest or lowest value, respectively, within a certain interval. These points are either critical points or endpoints of the interval.
To find these values:

• Evaluate the function at the critical points and endpoints of the domain.
• Compare these values to determine which is the highest (local maximum) and which is the lowest (local minimum).

For \( U(x) = x\sqrt{6-x} \), evaluate \( U(x) \) at \( x = 0 \), \( x = 4 \), and \( x = 6 \). After calculations, \( U(4) = 5.66 \) is a local maximum, while both \( x = 0 \) and \( x = 6 \) yield \( U(x) = 0 \) for local minima. Recognizing these helps in understanding the function’s behavior over its domain.