Math / Algebra

QuestionThe price $p$ (in dollars) and the quantity $x$ sold of a certain product satisfy the demand equation $x=-7 p+700$ . Answer parts (a) through (g). (い) villat pile $\mu$ maximizes ieverlue? $p=\$ 50$ (Simplify your answer. Type an integer or a decimal.) (d) What is the maximum revenue? $R=\$ 17500$ (Simplify your answer. Type an integer or a decimal.) (e) How many units are sold at this price? $x=350$ (Simplify your answer. Type an integer or a decimal.) (f) Graph R. Choose the correct graph below. A. B. C. D.

Studdy Solution

STEP 1

What is this asking? We're given a relationship between price and quantity sold, and we need to find the price that maximizes the total revenue, the maximum revenue itself, and the number of units sold at that price. Watch out! Don't mix up quantity sold ( $x$ ) with revenue ( $R$ ).
Revenue is the total money made, calculated as price times quantity.

STEP 2

1. Define the revenue function
2. Find the price that maximizes revenue
3. Calculate the maximum revenue
4. Calculate the number of units sold at the maximizing price

STEP 3

Alright, let's kick things off!
Revenue is simply the price per unit multiplied by the number of units sold.
We can write this mathematically as $R = p \cdot x$ .

STEP 4

We're given the relationship between $p$ and $x$ as $x = -7p + 700$ .
Let's substitute this expression for $x$ into our revenue equation.
This gives us $R(p) = p \cdot (-7p + 700)$ .

STEP 5

Now, let's expand this expression to get a nice, clean quadratic function for revenue: $R(p) = -7p^2 + 700p$ .
This tells us how the revenue changes as we change the price.

STEP 6

Our revenue function, $R(p) = -7p^2 + 700p$ , is a parabola that opens downwards (because the coefficient of $p^2$ is negative, $-7$ ).
This means it has a maximum value at its vertex.

STEP 7

The x-coordinate (in our case, the p-coordinate) of the vertex of a parabola in the form $ax^2 + bx + c$ is given by $-b/(2a)$ .
In our case, $a = -7$ and $b = 700$ , and $c=0$ .

STEP 8

So, the price that maximizes revenue is $p = -700 / (2 \cdot (-7))$ .
This simplifies to $p = -700 / (-14)$ , which gives us $p = \$$ 50.

STEP 9

Now that we know the maximizing price, $p = \$$ 50, let's plug it back into our revenue function, $R(p) = -7p^2 + 700p$ , to find the maximum revenue.

STEP 10

Substituting $p = 50$ , we get $R(50) = -7 \cdot (50)^2 + 700 \cdot 50$ .

STEP 11

Calculating this, we have $R(50) = -7 \cdot 2500 + 35000 = -17500 + 35000 = \$$ 17500.
That's our maximum revenue!

STEP 12

We found the price that maximizes revenue ( $p = \$$ 50).
Now, we need to find how many units are sold at this price.
We can use our original demand equation: $x = -7p + 700$ .

STEP 13

Substituting $p = 50$ , we get $x = -7 \cdot 50 + 700 = -350 + 700 =$ 350 units.

STEP 14

The price that maximizes revenue is $\$$ 50.
The maximum revenue is $\$$ 17500.
At this price, 350 units are sold.

Was this helpful?

Studdy Solution

STEP 1

STEP 2

1. Define the revenue function
2. Find the price that maximizes revenue
3. Calculate the maximum revenue
4. Calculate the number of units sold at the maximizing price

STEP 3

Alright, let's kick things off!
Revenue is simply the price per unit multiplied by the number of units sold.
We can write this mathematically as $R = p \cdot x$ .

STEP 4

We're given the relationship between $p$ and $x$ as $x = -7p + 700$ .
Let's substitute this expression for $x$ into our revenue equation.
This gives us $R(p) = p \cdot (-7p + 700)$ .

STEP 5

Now, let's expand this expression to get a nice, clean quadratic function for revenue: $R(p) = -7p^2 + 700p$ .
This tells us how the revenue changes as we change the price.

STEP 6

Our revenue function, $R(p) = -7p^2 + 700p$ , is a parabola that opens downwards (because the coefficient of $p^2$ is negative, $-7$ ).
This means it has a maximum value at its vertex.

STEP 7

The x-coordinate (in our case, the p-coordinate) of the vertex of a parabola in the form $ax^2 + bx + c$ is given by $-b/(2a)$ .
In our case, $a = -7$ and $b = 700$ , and $c=0$ .

STEP 8

So, the price that maximizes revenue is $p = -700 / (2 \cdot (-7))$ .
This simplifies to $p = -700 / (-14)$ , which gives us $p = \$$ 50.

STEP 9

Now that we know the maximizing price, $p = \$$ 50, let's plug it back into our revenue function, $R(p) = -7p^2 + 700p$ , to find the maximum revenue.

STEP 10

Substituting $p = 50$ , we get $R(50) = -7 \cdot (50)^2 + 700 \cdot 50$ .

STEP 11

Calculating this, we have $R(50) = -7 \cdot 2500 + 35000 = -17500 + 35000 = \$$ 17500.
That's our maximum revenue!

STEP 12

We found the price that maximizes revenue ( $p = \$$ 50).
Now, we need to find how many units are sold at this price.
We can use our original demand equation: $x = -7p + 700$ .

STEP 13

Substituting $p = 50$ , we get $x = -7 \cdot 50 + 700 = -350 + 700 =$ 350 units.

STEP 14

The price that maximizes revenue is $\$$ 50.
The maximum revenue is $\$$ 17500.
At this price, 350 units are sold.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Studdy Solution

STEP 1

STEP 2

1. Define the revenue function2. Find the price that maximizes revenue3. Calculate the maximum revenue4. Calculate the number of units sold at the maximizing price

STEP 3

Alright, let's kick things off!Revenue is simply the price per unit multiplied by the number of units sold.We can write this mathematically as R=p⋅xR = p \cdot xR=p⋅x.

STEP 4

We're given the relationship between ppp and xxx as x=−7p+700x = -7p + 700x=−7p+700.Let's **substitute** this expression for xxx into our revenue equation.This gives us R(p)=p⋅(−7p+700)R(p) = p \cdot (-7p + 700)R(p)=p⋅(−7p+700).

STEP 5

Now, let's **expand** this expression to get a nice, clean quadratic function for revenue: R(p)=−7p2+700pR(p) = -7p^2 + 700pR(p)=−7p2+700p.This tells us how the revenue changes as we change the price.

STEP 6

Our revenue function, R(p)=−7p2+700pR(p) = -7p^2 + 700pR(p)=−7p2+700p, is a parabola that opens downwards (because the coefficient of p2p^2p2 is negative, −7-7−7).This means it has a **maximum value** at its vertex.

STEP 7

The x-coordinate (in our case, the *p*-coordinate) of the vertex of a parabola in the form ax2+bx+cax^2 + bx + cax2+bx+c is given by −b/(2a)-b/(2a)−b/(2a).In our case, a=−7a = -7a=−7 and b=700b = 700b=700, and c=0c=0c=0.

STEP 8

So, the price that maximizes revenue is p=−700/(2⋅(−7))p = -700 / (2 \cdot (-7))p=−700/(2⋅(−7)).This simplifies to p=−700/(−14)p = -700 / (-14)p=−700/(−14), which gives us p=$p = \$p=$ **50**.

STEP 9

Now that we know the **maximizing price**, p=$p = \$p=$ **50**, let's plug it back into our revenue function, R(p)=−7p2+700pR(p) = -7p^2 + 700pR(p)=−7p2+700p, to find the **maximum revenue**.

STEP 10

Substituting p=50p = 50p=50, we get R(50)=−7⋅(50)2+700⋅50R(50) = -7 \cdot (50)^2 + 700 \cdot 50R(50)=−7⋅(50)2+700⋅50.

STEP 11

Calculating this, we have R(50)=−7⋅2500+35000=−17500+35000=$R(50) = -7 \cdot 2500 + 35000 = -17500 + 35000 = \$R(50)=−7⋅2500+35000=−17500+35000=$ **17500**.That's our maximum revenue!

STEP 12

We found the price that maximizes revenue (p=$p = \$p=$ **50**).Now, we need to find how many units are sold at this price.We can use our original demand equation: x=−7p+700x = -7p + 700x=−7p+700.

STEP 13

Substituting p=50p = 50p=50, we get x=−7⋅50+700=−350+700=x = -7 \cdot 50 + 700 = -350 + 700 = x=−7⋅50+700=−350+700= **350** units.

STEP 14

The price that maximizes revenue is $\$$ **50**.The maximum revenue is $\$$ **17500**.At this price, **350** units are sold.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. Define the revenue function
2. Find the price that maximizes revenue
3. Calculate the maximum revenue
4. Calculate the number of units sold at the maximizing price

Alright, let's kick things off!
Revenue is simply the price per unit multiplied by the number of units sold.
We can write this mathematically as $R = p \cdot x$ .

We're given the relationship between $p$ and $x$ as $x = -7p + 700$ .
Let's substitute this expression for $x$ into our revenue equation.
This gives us $R(p) = p \cdot (-7p + 700)$ .

Now, let's expand this expression to get a nice, clean quadratic function for revenue: $R(p) = -7p^2 + 700p$ .
This tells us how the revenue changes as we change the price.

Our revenue function, $R(p) = -7p^2 + 700p$ , is a parabola that opens downwards (because the coefficient of $p^2$ is negative, $-7$ ).
This means it has a maximum value at its vertex.

The x-coordinate (in our case, the p-coordinate) of the vertex of a parabola in the form $ax^2 + bx + c$ is given by $-b/(2a)$ .
In our case, $a = -7$ and $b = 700$ , and $c=0$ .

So, the price that maximizes revenue is $p = -700 / (2 \cdot (-7))$ .
This simplifies to $p = -700 / (-14)$ , which gives us $p = \$$ 50.

Now that we know the maximizing price, $p = \$$ 50, let's plug it back into our revenue function, $R(p) = -7p^2 + 700p$ , to find the maximum revenue.

Substituting $p = 50$ , we get $R(50) = -7 \cdot (50)^2 + 700 \cdot 50$ .

Calculating this, we have $R(50) = -7 \cdot 2500 + 35000 = -17500 + 35000 = \$$ 17500.
That's our maximum revenue!

We found the price that maximizes revenue ( $p = \$$ 50).
Now, we need to find how many units are sold at this price.
We can use our original demand equation: $x = -7p + 700$ .

Substituting $p = 50$ , we get $x = -7 \cdot 50 + 700 = -350 + 700 =$ 350 units.

The price that maximizes revenue is $\$$ 50.
The maximum revenue is $\$$ 17500.
At this price, 350 units are sold.