Smart TV Sales Analysis: Demand, Revenue, And Profit Maximization
Hey guys! Let's dive into a cool math problem about a manufacturer selling smart TVs. We're going to explore how the price of the TVs affects how many they sell, how much money they make (revenue), and ultimately, how much profit they bring in. Buckle up, because we're about to calculate some cool stuff!
Understanding the Demand Function
Okay, so here's the deal: a manufacturer is currently selling 1050 smart TVs each week at a price of $540 per TV. But here's the interesting part – they've done a market survey, and it shows that if they offer a rebate (a discount) to buyers, they'll sell more TVs. Specifically, for every $10 rebate they offer, they'll sell 100 more TVs each week. This relationship between the price and the quantity sold is what we call the demand function. Our task is to find a function that will tell us the relationship between price and quantity, represented by p(x).
Let's break it down: we know that the initial price is $540, and the initial quantity sold is 1050 TVs. We can represent the quantity of TVs sold (x) as a function of the rebate offered. Let 'r' be the amount of rebate offered. Because the number of TV increases by 100 for every $10 rebate, we can show that for every $10 rebate the quantity increases by 100. So, we can create an equation that expresses the quantity of TV sold based on rebate, which is x = 1050 + 100 * (r / 10), where r is the rebate value. So we can say r = 10 * ((x - 1050) / 100).
The key to finding the demand function, p(x), is understanding how the price changes with the number of TVs sold. Let's think step by step, the price changes based on the rebate. As the number of TVs sold (x) changes, the rebate offered is determined as well. The rebate (r) is the amount discounted from the original price. Thus, the price should be:
p(x) = 540 - r.
We already know that r = 10 * ((x - 1050) / 100), we can replace r, and we get the demand function as follows:
p(x) = 540 - 10 * ((x - 1050) / 100)
Simplifying this, we get:
p(x) = 540 - 0.1(x - 1050)
p(x) = 540 - 0.1x + 105
p(x) = 645 - 0.1x
This demand function, p(x) = 645 - 0.1x, tells us the price the manufacturer needs to charge to sell a specific number of TVs. For example, if they want to sell 1150 TVs (x = 1150), they would need to charge a price of p(1150) = 645 - 0.1(1150) = $530 per TV. So that is the demand function.
Diving into Revenue and Revenue Maximization
Alright, now that we have the demand function, let's talk about revenue. Revenue is simply the total amount of money the manufacturer brings in from selling TVs. It's calculated by multiplying the price per TV by the number of TVs sold. Mathematically, Revenue (R(x)) = price (p(x)) * quantity (x). Using the demand function we just calculated, we can determine the revenue function as:
R(x) = (645 - 0.1x) * x
R(x) = 645x - 0.1x^2
So that is the revenue function. Now, the manufacturer wants to know how to maximize their revenue. How can they set the price to earn as much money as possible? This is where calculus comes in handy! To find the maximum revenue, we need to find the critical points of the revenue function. Critical points are the points where the derivative of the function is equal to zero or undefined. The derivative gives us the rate of change of the revenue with respect to the number of TVs sold. Let's find the derivative of the revenue function R(x):
R'(x) = 645 - 0.2x
To find the critical point, we set the derivative equal to zero and solve for x:
0 = 645 - 0.2x
0.2x = 645
x = 3225
So, the critical point is x = 3225. This means that to maximize revenue, the manufacturer should aim to sell 3225 TVs. But wait! We need to make sure this critical point is a maximum, not a minimum. We can use the second derivative test for this. Let's find the second derivative of the revenue function:
R''(x) = -0.2
Since the second derivative is negative (-0.2), this indicates that the critical point is indeed a maximum. Now we can calculate the maximum revenue by plugging this value of x back into the revenue function:
R(3225) = 645(3225) - 0.1(3225)^2
R(3225) = 2080125 - 103922.25
R(3225) = 1040906.25
So, the maximum revenue is $1040906.25. To achieve this maximum revenue, we need to find the price that corresponds to selling 3225 TVs. We can use the demand function again:
p(3225) = 645 - 0.1(3225)
p(3225) = 645 - 322.5
p(3225) = 322.5
Therefore, the manufacturer should set the price at $322.5 to sell 3225 TVs and maximize their revenue, which is $1040906.25. That's a lot of money!
Unleashing Profit and Profit Maximization
Now, let's explore profit. Profit is what the manufacturer makes after subtracting the cost of producing the TVs from their revenue. The problem doesn't give us any information on costs. We would need the cost function, C(x), which is the total cost of producing x TVs. Let's assume that the cost function is given as C(x) = 100000 + 150x. This means the manufacturer has fixed costs of $100,000 (like rent and equipment) and variable costs of $150 per TV (like materials and labor).
Now, let's calculate the profit function, P(x). The profit is Revenue - Cost.
P(x) = R(x) - C(x)
P(x) = (645x - 0.1x^2) - (100000 + 150x)
P(x) = 645x - 0.1x^2 - 100000 - 150x
P(x) = -0.1x^2 + 495x - 100000
So the profit function is P(x) = -0.1x^2 + 495x - 100000. To maximize profit, we'll again find the critical points by taking the derivative of the profit function and setting it to zero:
P'(x) = -0.2x + 495
Set the derivative equal to zero:
0 = -0.2x + 495
0.2x = 495
x = 2475
So, the critical point is x = 2475. This means that to maximize profit, the manufacturer should aim to sell 2475 TVs. Now let's check this is a maximum with the second derivative test.
P''(x) = -0.2
The second derivative is negative (-0.2), confirming that the critical point is indeed a maximum. Now we can calculate the maximum profit by plugging this value of x back into the profit function:
P(2475) = -0.1(2475)^2 + 495(2475) - 100000
P(2475) = -612562.5 + 1224875 - 100000
P(2475) = 512312.5
So, the maximum profit is $512312.5. We also need to find the price that corresponds to selling 2475 TVs to maximize the profit. Using the demand function:
p(2475) = 645 - 0.1(2475)
p(2475) = 645 - 247.5
p(2475) = 397.5
Therefore, the manufacturer should set the price at $397.5 to sell 2475 TVs and maximize the profit, which is $512312.5. By using the demand function, we can calculate the price to maximize profits. Pretty sweet!
Conclusion
In a nutshell, we've gone from analyzing the initial sales figures to discovering how the manufacturer can optimize their pricing strategy to maximize revenue and profit. We started with the demand function and then used calculus to find the optimal selling price and the quantity of TVs to sell. This process not only helps the manufacturer make more money but also helps them understand the market dynamics and how price changes impact sales. Understanding the demand, revenue, and profit is very useful for any business. Keep in mind that these calculations are based on the assumptions we made, such as a linear demand function and a specific cost function. But the principles remain the same: understand your market, calculate your costs and revenue, and use math to make smart business decisions!
