Question

The price, \(p,\) in dollars, set by a manufacturer for \(x\) tonnes of steel is \(p(x)=12 x\left(\frac{x+2}{x+1}\right) .\) Using the quotient of functions, determine whether the price per tonne decreases as the number of tonnes increases, algebraically and graphically.

Short answer

The price per tonne initially increases, then decreases after \( x = \sqrt{2} - 1 \).

Step-by-step solution

Understand the Function

Identify and understand the given price function. The function is given as follows: \[ p(x) = 12x \left( \frac{x+2}{x+1} \right) \]

Simplify the Function

Simplify the given function by performing algebraic operations: \[ p(x) = 12x \left( \frac{x+2}{x+1} \right) = 12 \left( x + 2 - \frac{2}{x+1} \right) = 12 \left( 1 + \frac{1}{x+1} \right) \]This simplifies to: \[ p(x) = 12x + \frac{24}{x+1} \]

Find the Derivative of the Price Function

To determine if the price per tonne decreases, find the derivative of the price function. Use the quotient rule for derivatives: \[ p'(x) = \frac{d}{dx} \left( 12x + \frac{24}{x+1} \right) \]Applying the quotient rule and chain rule: \[ p'(x) = 12 - \frac{24}{(x+1)^2} \]

Analyze the Derivative

Solve for the critical points by setting \( p'(x) = 0 \): \[ 12 - \frac{24}{(x+1)^2} = 0 \]Solve for \( x \): \[ 12 = \frac{24}{(x+1)^2} \]\[ (x+1)^2 = 2 \]\[ x+1 = \sqrt{2} \]\[ x = \sqrt{2} - 1 \]Therefore, the critical point is \( x = \sqrt{2} - 1 \).

Evaluate the Derivative's Sign

Determine the sign of \( p'(x) \) for intervals separated by the critical point \( x = \sqrt{2} - 1 \). For \( x < \sqrt{2} - 1 \) and \( x > \sqrt{2} - 1 \): - If \( x < \sqrt{2} - 1 \), \( p'(x) > 0 \) - If \( x > \sqrt{2} - 1 \), \( p'(x) < 0 \)This means that price per tonne initially increases, then decreases after a certain point.

Graphical Analysis

Plot the function \( p(x) = 12 x \left( \frac{x+2}{x+1} \right) \) on a graph to visually determine the behavior as \( x \) increases. Observe the curve to identify the regions where the function's slope is positive and where it is negative, affirming the results from the derivative analysis.

Key concepts

quotient rule

The quotient rule is essential for finding the derivative of a function that is the ratio of two differentiable functions. The rule states that for a function expressed as a quotient of two functions, say \( f(x) = \frac{u(x)}{v(x)} \), its derivative is given by:

\[ f'(x) = \frac{u'(x) \cdot v(x) - u(x) \cdot v'(x)}{v(x)^2} \]
To apply this rule to our function \( p(x) = 12x \cdot \frac{x+2}{x+1} \), we need to differentiate both the numerator (u(x) = 12x(x + 2)) and the denominator (v(x) = x + 1). The derivative of the numerator using the product rule is:

\[ u'(x) = 12 \cdot (x+2) + 12x \cdot 1 = 12x + 24 \]
The derivative of the denominator is:

\[ v'(x) = 1 \]
Substituting these into the quotient rule gives us the derivative we need to analyze how the price per tonne changes with the increase in the number of tonnes.

derivative analysis

Derivative analysis involves finding the derivative of the function and understanding what it tells us about the function's behavior. For our function \( p(x) = 12x \frac{x+2}{x+1} \), the derivative we found is:

\[ p'(x) = 12 - \frac{24}{(x+1)^2} \]
By setting this derivative equal to zero, we find the critical points. These are the points where the slope of the tangent to the curve is zero, indicating possible local maxima, minima, or inflection points.

Solving: \[ 12 - \frac{24}{(x+1)^2} = 0 \]
we get: \[ 12(x+1)^2 = 24 \]
or: \[ (x+1)^2 = 2 \]
which simplifies to: \[ x = \sqrt{2} - 1 \].
This critical point helps us decide the intervals to test the sign of \( p'(x) \) to determine if the function is increasing or decreasing in those intervals.

graphical analysis

Graphical analysis helps visualize the behavior of the function \( p(x) = 12x \frac{x+2}{x+1} \). By plotting the function, we can observe where the function increases or decreases and confirm the results of our derivative analysis.

When you plot this function, you typically observe that it initially rises to the critical point \( x = \sqrt{2} - 1 \) and then begins to fall. The slope of the curve changes from positive (increasing) to negative (decreasing) at this point. This confirms the critical point analysis from the derivative calculation.
Visualizing the function graph makes it simpler to understand the effect of changes in \( x \) on the price per tonne, providing a clear picture of the turning points and overall trend.

critical points

Critical points occur where the derivative of a function is zero or undefined. These points are significant because they help identify where the function changes its behavior. For our function \( p(x) = 12x \frac{x+2}{x+1} \), setting the derivative equal to zero gave us the critical point:

\[ x = \sqrt{2} - 1 \]
To determine whether this critical point represents a maximum or minimum, we can analyze the sign of the derivative around \( x = \sqrt{2} - 1 \).
- For \( x < \sqrt{2} - 1 \), we find that \( p'(x) > 0 \), indicating the function is increasing.
- For \( x > \sqrt{2} - 1 \), we find that \( p'(x) < 0 \), indicating the function is decreasing.
Thus, the critical point at \( x = \sqrt{2} - 1 \) represents a local maximum. The behavior at and around the critical points helps us understand the nature of the function's overall trend, in this case showing an initial increase in price per tonne followed by a decrease beyond the critical point.