Problem 73 - Calculus and its applications

Question

The Candy Factory sells candy by the pound, charging \(\$ 1.50\) per pound for quantities up to and including 20 pounds. Above 20 pounds, the Candy Factory charges \(\$ 1.25\) per pound for the entire quantity. If \(x\) represents the number of pounds, the price function is $$ p(x)=\left\\{\begin{array}{ll} 1.50 x, & \text { for } x \leq 20 \\ 1.25 x, & \text { for } x>20 \end{array}\right. $$ Find \(\lim _{x \rightarrow 2^{-}} p(x), \lim _{x \rightarrow 20^{+}} p(x),\) and \(\lim _{x \rightarrow 20} p(x)\).

Step-by-Step Solution

Answer

The limits are \( \lim_{x \to 2^-} p(x) = 3 \), \( \lim_{x \to 20^+} p(x) = 25 \), and \( \lim_{x \to 20} p(x) \) does not exist.

Step 1: Understanding the Problem

We are given a piecewise price function based on the weight of the candy, where the price per pound is \(1.50 if the weight is 20 pounds or less and \)1.25 if the weight is more than 20 pounds. We need to find the limits of this function as \( x \) approaches certain points.

Step 2: Calculating \( \lim_{x \to 2^-} p(x) \)

Since \( x = 2 \) is less than 20, we use the first case of the piecewise function: \( p(x) = 1.50x \). As \( x \to 2^- \), the expression becomes \( 1.50 \times 2 \), which equals 3. Thus, \( \lim_{x \to 2^-} p(x) = 3 \).

Step 3: Calculating \( \lim_{x \to 20^+} p(x) \)

For values greater than 20, we use the second case of the piecewise function: \( p(x) = 1.25x \). As \( x \to 20^+ \), the expression becomes \( 1.25 \times 20 \), which equals 25. Therefore, \( \lim_{x \to 20^+} p(x) = 25 \).

Step 4: Calculating \( \lim_{x \to 20} p(x) \)

We need to consider both one-sided limits. We already found that \( \lim_{x \to 20^-} p(x) = 1.50 \times 20 = 30 \) and \( \lim_{x \to 20^+} p(x) = 25 \) from the previous steps. Since these one-sided limits are not equal, \( \lim_{x \to 20} p(x) \) does not exist.

Key Concepts

Piecewise Functions

Piecewise functions are like multiple functions in one. They switch their expressions based on different intervals of the input value. In the scenario given, we have the Candy Factory's pricing model. Here, the cost per pound changes depending on whether you buy up to 20 pounds or more than 20 pounds. Let's look closely.

• The price is \\(1.50 per pound when you buy 20 pounds or less.
• It's a bit cheaper, \\)1.25 per pound, when you purchase more than 20 pounds.

The function can be written as two different formulas, each defined over specific portions of the input space. This is what makes it 'piecewise'. Such functions are handy in modeling situations where the rules change at certain points. They're widely used in various fields, from economics to physics.

One-Sided Limits

One-sided limits help us understand how a function behaves as it approaches a specific point from only one side, either from the left or right. This is especially crucial in piecewise functions where changes occur at certain breakpoints.
For the Candy Factory's price function, consider the left-hand limit as you approach a smaller quantity, like \(x = 2^{-}\), which simply means you're getting very close to 2 from the left side. For \(x < 20\), the first part of the piecewise function applies: \(p(x) = 1.50x\).
As you near \(x = 2\), the value of the function becomes \(1.50 \times 2 = 3\).Similarly, the right-hand limit, such as \(x = 20^{+}\), deals with how the function behaves as we move past 20 to the right. Beyond 20 pounds, we use \(p(x) = 1.25x\), leading to \(1.25 \times 20 = 25\).
These clear but distinct ways the function addresses limits from different sides are fundamental in analyzing discontinuities and understanding real-world constraints.

Limit Calculation

Calculating limits in piecewise functions involves examining smooth transitions or discontinuities at switching points.
For example, let us use the Candy Factory pricing to calculate \(\lim_{x \to 20} p(x)\). We need to determine what happens when we approach 20 from both sides.First, approach 20 from the left, applying \(x \leq 20\): \(\lim_{x \to 20^{-}} p(x) = 1.50 \times 20 = 30\).
Next, approach from the right (\(x > 20\): \(\lim_{x \to 20^{+}} p(x) = 1.25 \times 20 = 25\).We observe that approaching 20 from the left yields 30, while from the right yields 25. Since these values do not match, the overall limit \(\lim_{x \to 20} p(x)\) does not exist. This reflects a discontinuity at 20 in this piecewise function. Limit calculations like these are essential for identifying and understanding the nature of function behaviors at critical points.