A company manufactures and sells x smartphones per week. The...

A company manufactures and sells $x$ smartphones per week. The weekly price-demand and cost equations are, respectively, $$ p=500-0.5 x \quad \text { and } \quad C(x)=20,000+135 x $$ (A) What price should the company charge for the phones, and how many phones should be produced to maximize the weekly revenue? What is the maximum weekly revenue? (B) What is the maximum weekly profit? How much should the company charge for the phones, and how many phones should be produced to realize the maximum weekly profit?

A company manufactures and sells $x$ smartphones per week. The weekly price-demand and cost equations are, respectively,
$$
p=500-0.5 x \quad \text { and } \quad C(x)=20,000+135 x
$$
(A) What price should the company charge for the phones, and how many phones should be produced to maximize the weekly revenue? What is the maximum weekly revenue?
(B) What is the maximum weekly profit? How much should the company charge for the phones, and how many phones should be produced to realize the maximum weekly profit?

Key Concepts

Price-Demand Relationship

This concept defines the relationship between the price of a product and the quantity demanded by consumers. It typically reflects that as the price decreases, the demand increases, and vice versa. Understanding this relationship is essential in setting optimal pricing strategies in order to meet market demand conditions and maximize sales or profits.

Revenue Function

The revenue function represents the total income generated from the sale of goods or services and is generally formulated as the product of price and quantity sold. In problems involving price-demand equations, the revenue function becomes a function of quantity, which allows for analysis and optimization of sales revenue with respect to changes in production or pricing levels.

Quadratic Optimization

Many revenue and profit functions are quadratic in nature, meaning they can be represented as parabolic graphs. Quadratic optimization involves finding the vertex of the parabola, which corresponds to the maximum or minimum value of the function. This process is crucial in maximizing revenue or profit when the function exhibits a concave shape (for maximization problems) or a convex shape (for minimization problems).

Cost Function

The cost function describes the total cost of producing a certain number of goods, often incorporating both fixed and variable costs. Fixed costs are incurred regardless of production level, while variable costs depend on the quantity produced. Understanding the cost function enables businesses to evaluate production efficiency and make informed decisions regarding pricing and output levels.

Profit Function

The profit function is defined as the difference between total revenue and total cost. It provides a measure of the financial gain resulting from business operations and is a key indicator for decision-making. By analyzing the profit function, companies can determine the optimal production level and pricing strategy to maximize profitability, balancing revenue gains against production expenses.

Transcript

00:01 For this problem, we are told that a company manufactures and sells x smartphones per week.

00:06 The weekly price demand and cost equations are p equals 500 minus 0 .5x and c of x equals 20 ,000 plus 135x.

00:14 For the first part, we are asked, what price should the company charge for the phones and how many phones should be produced to maximize the weekly revenue and what is the maximum weekly revenue? then for part b, we are asked, what is the maximum weekly profit and how much should the company charge? for the phones, how many phones should be produced to realize the maximum weekly profit.

00:37 Keep in mind, profit versus revenue.

00:40 So, for part a regarding the price, and specifically the weekly revenue, we know that the revenue is going to equal the price times the number of units sold.

00:54 So our revenue is going to actually equal x times 500 minus 0 .5x.

01:05 Which then means we can expand this out as negative 0 .5x squared plus 500x.

01:14 One second here.

01:15 Now to maximize the weekly revenue, we want to take the derivative of the revenue with respect to the number of units sold.

01:22 So that is going to give us, so we'll have negative 0 .5 times 2, so that's just going to be negative 1x plus 500.

01:32 So we want to figure out how many units sold will bring this to 0.

01:37 Obviously that is going to be just when x equals 500.

01:42 We'll have that the derivative of the revenue with respect to units sold is going to be zero.

01:49 And we can tell, you know, this is a parabola.

01:53 So up here is going to be the maximum.

01:55 That's the turning point where the derivative is zero.

01:58 So to figure out what the maximum weekly revenue then will be, we just plug that back in...

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