The price p (in dollars) and the quantity x sold of a...

The price $p$ (in dollars) and the quantity $x$ sold of a certain product obey the demand equation $$x=-5 p+100 \quad 0<p \leq 20$$ (a) Express the revenue $R$ as a function of $x$. (b) What is the revenue if 15 units are sold? (c) What quantity $x$ maximizes revenue? What is the maximum revenue? (d) What price should the company charge to maximize revenue? (e) What price should the company charge to earn at least $\$ 480$ in revenue?

The price $p$ (in dollars) and the quantity $x$ sold of a certain product obey the demand equation $$x=-5 p+100 \quad 0<p \leq 20$$
(a) Express the revenue $R$ as a function of $x$.
(b) What is the revenue if 15 units are sold?
(c) What quantity $x$ maximizes revenue? What is the maximum revenue?
(d) What price should the company charge to maximize revenue?
(e) What price should the company charge to earn at least $\$ 480$ in revenue?

Key Concepts

Domain Constraints

Domain constraints refer to the limitations set on the variables involved, such as restrictions on price or quantity. These constraints ensure that the solutions obtained from mathematical models are practical and economically feasible within the context of the market environment.

Optimization Techniques

Optimization techniques involve finding the maximum or minimum values of a function, which, in the context of a quadratic revenue function, is achieved by identifying the vertex of the parabola. This approach helps determine the optimal quantity sold or price charged to maximize revenue.

Demand Equation

A demand equation models the relationship between the price of a product and the quantity sold, showing how changes in price affect consumer demand. It is fundamental in economics for analyzing market behavior and helps in determining pricing strategies to optimize revenue.

Revenue Function

The revenue function is derived by multiplying the price by the quantity sold. This function is essential in business and economics as it relates sales volume and pricing strategy to the overall income generated, allowing firms to evaluate the effects of price adjustments on their earnings.

Quadratic Functions

Quadratic functions are polynomial functions of degree two that typically form a parabolic shape when graphed. In revenue analysis, the revenue function can become quadratic when the demand equation is linear, leading to a parabolic curve that has a maximum or minimum, depending on the orientation of the parabola.

Transcript

00:01 So in this problem, we're given that the price p and the quantity x sold of a certain product, all of this relationship, x equals minus 5p plus 100, where the price is greater than zero, but less than 20.

00:31 And we're first asked to express the revenue, r, as a function of x.

00:42 Okay, so revenue remember, this is price times quantity.

00:55 So i'm going to get the price as a function of x, and then the quantity, of course, is x.

01:01 And so solve this equation for p.

01:06 So that's x minus 100, x minus 100, and then divide all that by minus 5.

01:18 Multiple of that times x, the quantity, right? so that means i have minus 1 5th x squared, because this is minus 1 5th x for the first term here.

01:33 And then minus 100 over minus 5, that's positive 20, right? so it'd be plus 20x.

01:42 So there is our revenue equation, revenue function, as a function of x, the quantity.

01:51 Okay, next we're asked, what is the revenue if 15 units are sold? so we need r of 15.

01:58 Which means we have minus 1 5th times 15 squared plus 20 times 15.

02:07 Okay.

02:08 Well, 15 squared is 225.

02:14 So this is minus a fifth times 225 plus 20 times 300.

02:27 And 225 divided by 5 is 45.

02:32 This is minus 45 plus 300, which is $255.

02:45 Okay.

02:50 Next, we're asked what quantity x maximizes revenue and what is the maximum revenue.

02:55 So a couple ways we can do this, right? we could take the derivative of this, set it equal to zero.

03:03 Derivative of that, set it equal to zero, and find the maximum price, or i'm sorry, maximum quantity.

03:09 And then calculate the price.

03:12 Do that.

03:13 The other way to do it would be to graph it and see where the maximum point is.

03:18 So let's do the derivative first.

03:20 So when i do the derivative, i bring the two exponent down front, so that's minus 2 over 5 times x and to the power of 2 minus 1 plus 20.

03:42 Because the x has a 1 power, bring the 1 down, that's 20 times 1 is 1.

03:46 And the exponent is 1 minus 1 is 0, so that's x is just 1.

03:52 Okay, set this equal to 0 now.

03:57 And so what happens? i have minus 2 over 5x equals minus 20, multiply both sides by negative 5 over 2.

04:08 So x is negative 20 times negative 5 over 2.

04:13 Negative times negative is positive.

04:16 2 goes into 20, 10 times.

04:19 So this quantity is 50.

04:21 There's my quantity and price remember was negative one -fifth x plus 20.

04:40 So that's negative one -fifth times 50 plus 20.

04:47 Well, five goes into 50 ten times...

Please give Ace some feedback