Maximizing Revenue: Sicard Sports Watch Pricing

Hey everyone, let's dive into the awesome world of maximizing revenue! We're going to explore how the Sicard sports watch can optimize its pricing strategy to make some serious cash. This is where the math gets fun, guys. We'll be using the equation ${p = \frac{45}{0.01x^2 + 1}}$ where ${p}$ represents the unit price in dollars, and ${x}$ represents the quantity demanded in units of a thousand (with ${0 \leq x \leq 20}$). Think of this as a real-world scenario where you're trying to figure out the best price to sell your cool watches.

Understanding the Demand Equation and Revenue

So, the equation ${p = \frac{45}{0.01x^2 + 1}}$ is super important because it tells us the relationship between the price of the watch and how many people want to buy it. It's called the demand equation. As the price goes up, the quantity demanded goes down, and vice versa. It's like a seesaw, you know? Now, the revenue (R) is the total amount of money the company brings in. We calculate it by multiplying the price (p) by the quantity sold (x, in thousands). That is, ${R = px}$. Since we have the demand equation, we can rewrite the revenue equation using it. So, we'll get ${R(x) = x \cdot \frac{45}{0.01x^2 + 1} = \frac{45x}{0.01x^2 + 1}}$. The revenue is now a function of ${x}$, which is the number of watches sold (in thousands). Our mission, should we choose to accept it, is to figure out the optimal value of ${x}$ to maximize the revenue. You know, what quantity sold will bring in the most money? Before we go any further, it's very important to note that the quantity demanded each month of the Sicard sports watch is related to the unit price by the equation, where ${p}$ is measured in dollars and ${x}$ is measured in units of a thousand. It is also important to consider the interval of this equation because it affects the overall outcome of our calculations. The interval here is between 0 and 20. The interval is also a very important factor, because we can only optimize the equation within the confines of our interval. Otherwise, our calculations will be false.

Now, we'll need to use some calculus magic to find the maximum revenue. This involves finding the derivative of the revenue function, setting it equal to zero, and solving for ${x}$. That ${x}$ value will give us the quantity that maximizes revenue. The fun thing about this is that, as we change the value of ${x}$, the revenue will also change. It will increase or decrease based on the values that we put in. However, our main goal here is to find the maximum value, and in order to do that, we must take the derivative and set the function to 0. We will get an x value that can help us. The maximum revenue is the most we can get from the equation. In order to get the maximum value, we must know the exact value of ${x}$ in the demand equation. If we put in the values of ${x}$ in the demand equation we can get the value of ${p}$, and then multiply these two values to find the revenue. The revenue value we obtain will give us our maximum value. Keep in mind that, as the values of ${x}$ changes, the values of ${p}$ and the revenue will also change. Now, let's go on to the next part and find the derivative!

Finding the Optimal Quantity Using Calculus

Okay, buckle up, because here comes the calculus! We have our revenue function ${R(x) = \frac{45x}{0.01x^2 + 1}}$, and now we need to find its derivative, ${R'(x)}$. We can do this using the quotient rule, which states that if ${f(x) = \frac{u(x)}{v(x)}}$, then ${f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}}$. Applying the quotient rule to our revenue function, where ${u(x) = 45x}$ and ${v(x) = 0.01x^2 + 1}$, we get the derivative. So, ${u'(x) = 45}$ and ${v'(x) = 0.02x}$. Plugging these into the quotient rule gives us ${R'(x) = \frac{45(0.01x^2 + 1) - 45x(0.02x)}{(0.01x^2 + 1)^2}}$. Simplifying this, we get ${R'(x) = \frac{45(0.01x^2 + 1 - 0.02x^2)}{(0.01x^2 + 1)^2} = \frac{45(1 - 0.01x^2)}{(0.01x^2 + 1)^2}}$. The derivative gives us the rate of change of the revenue with respect to the quantity. To find the critical points (potential maximums or minimums), we need to set the derivative equal to zero and solve for ${x}$. So, we have ${0 = \frac{45(1 - 0.01x^2)}{(0.01x^2 + 1)^2}}$. This means ${1 - 0.01x^2 = 0}$, and solving for ${x}$ gives us ${x^2 = \frac{1}{0.01} = 100}$, so ${x = \pm 10}$. But remember, ${x}$ represents the quantity of watches sold (in thousands), and we can't sell a negative quantity. So, we'll take the positive value, ${x = 10}$. This is the critical point. Since this is our only critical point within the interval ${[0, 20]}$, it's highly likely this is where we'll find the maximum revenue. Now, we must consider the second derivative to ensure that this value is a maximum.

To ensure that this is the maximum revenue, we need to apply the second derivative test. Let's find the second derivative of ${R(x)}$, ${R''(x)}$. It is more complex, but here's the result: ${R''(x) = \frac{-0.9x(0.01x^2 + 1)^2 - 45(1 - 0.01x^2)(2)(0.01x^2 + 1)(0.02x)}{(0.01x^2 + 1)^4}}$. If we evaluate ${R''(10)}$, we will get a negative value, confirming that ${x = 10}$ is indeed a maximum. This is because the second derivative being negative indicates that the function is concave down at that point, which means it's a local maximum. Now that we know ${x = 10}$ maximizes revenue, we can proceed to find the price and the maximum revenue.

Determining the Optimal Price and Maximum Revenue

Alright, we've found that the optimal quantity to maximize revenue is ${x = 10}$ (thousand watches). Now, let's figure out what price we should set to sell that quantity and what the maximum revenue will be. We'll go back to our demand equation, ${p = \frac{45}{0.01x^2 + 1}}$. Plugging in ${x = 10}$, we get ${p = \frac{45}{0.01(10)^2 + 1} = \frac{45}{0.01(100) + 1} = \frac{45}{1 + 1} = \frac{45}{2} = 22.5}$. So, the optimal price is $22.50 per watch. Great! Now, to calculate the maximum revenue, we'll use the revenue equation ${R(x) = px}$. We know ${x = 10}$ and ${p = 22.5}$, so the maximum revenue is ${R = 22.5 \cdot 10 = 225}$. But remember that ${x}$ is in thousands, so this is $225,000. So, by selling 10,000 watches at $22.50 each, the Sicard sports watch company will achieve a maximum revenue of $225,000. Not bad at all, right?

Keep in mind that this is based on the demand equation given. In the real world, the demand equation might change based on factors like advertising, competition, and changes in consumer preferences. But, using calculus, we've successfully found the optimal pricing strategy to maximize revenue based on the given information. This is a powerful demonstration of how math can be used to make smart business decisions. Also, the same technique can be used with different kinds of equations. This can also be used in different companies with different demands and equations. The important thing is to understand the equations and apply calculus! This also shows us how much the value of math can be when applied to real life situations.

Summarizing the Optimization Process

To wrap things up, let's quickly recap the steps we took to maximize revenue:

1. Understand the Demand Equation: We started with the demand equation, which describes the relationship between price and quantity demanded.
2. Define the Revenue Function: We created a revenue function by multiplying the price by the quantity sold, expressed as a function of ${x}$.
3. Find the Derivative: We took the derivative of the revenue function to find the rate of change of revenue.
4. Find Critical Points: We set the derivative equal to zero and solved for ${x}$ to find the critical points. These are potential maximums or minimums.
5. Use the Second Derivative Test: We employed the second derivative test to confirm that the critical point was a maximum.
6. Determine the Optimal Price: We plugged the optimal quantity ${x}$ into the demand equation to find the corresponding price.
7. Calculate Maximum Revenue: We multiplied the optimal price by the optimal quantity to find the maximum revenue.

This process is crucial for businesses. It allows them to make informed decisions about pricing and production levels, which can lead to increased profitability and market success. Understanding these concepts can be a game-changer. So, keep up the great work, everyone. If you have any questions, feel free to ask! This is a great skill that can be applied to different aspects of life, not just business! It also shows us how valuable the study of math can be. So, keep it up!

Also, remember that real-world business scenarios can be more complex than this simplified model. However, the core principles remain the same. The use of the derivative to find the maximum values will always come in handy! This process can be applied to many different kinds of equations and businesses. The important thing is to understand the concepts and how to apply them. That's all for now, guys. Keep learning, and keep maximizing your potential!