Math / Algebra

QuestionThe weekly revenue from the sale of $x$ units of a service is given by $R(x)=110 x-5 x^{2}$ thousand dollars, where $0 \leq x \leq 19$ . How many units should be sold to maximize the revenue?

Studdy Solution

STEP 1

What is this asking? We're trying to find the sweet spot of how many units of a service we need to sell to make the most money each week. Watch out! Don't mix up revenue with profit!
Revenue is just the total money coming in, not accounting for costs.

STEP 2

1. Define the function
2. Find the derivative
3. Set the derivative to zero
4. Solve for $x$
5. Check the boundaries

STEP 3

Our revenue function, $R(x)$ , tells us how much money we make each week based on the number of units, $x$ , we sell.
It's given by $R(x) = 110x - 5x^2$ , where $R(x)$ is in thousands of dollars.
Remember, $x$ has to be between 0 and 19 inclusive!

STEP 4

To find where our revenue is maximized, we need to find the derivative of our revenue function, $R'(x)$ .
The derivative tells us how much the revenue is changing for each additional unit sold.

STEP 5

Using the power rule, the derivative of $110x$ is $110$ , and the derivative of $-5x^2$ is $-10x$ .
So, our derivative is $R'(x) = 110 - 10x$ .

STEP 6

When the derivative is zero, it means the revenue isn't increasing or decreasing – it's at a peak (or a valley)!
We want the peak, the maximum revenue.
So, we set $R'(x) = 0$ .
That gives us $110 - 10x = 0$ .

STEP 7

Now, let's solve for $x$ !
We can add $10x$ to both sides of the equation: $110 = 10x$ .

STEP 8

Then, divide both sides by 10 to get $x = 11$ .
This tells us that selling 11 units might give us the maximum revenue.

STEP 9

Remember how $x$ has to be between 0 and 19?
We need to check the revenue at these boundaries, just in case they give us higher revenue than selling 11 units.

STEP 10

If $x = 0$ , $R(0) = 110 \cdot 0 - 5 \cdot 0^2 = 0$ .
Well, selling zero units gives us zero revenue, which makes sense!

STEP 11

If $x = 19$ , $R(19) = 110 \cdot 19 - 5 \cdot 19^2 = 2090 - 1805 = 285$ .
That's $\$285,000$ !

STEP 12

Now, let's check our potential maximum at $x = 11$ : $R(11) = 110 \cdot 11 - 5 \cdot 11^2 = 1210 - 605 = 605$ .
That's $\$605,000$ !

STEP 13

Since $\$605,000$ is greater than both $\$0$ and $\$285,000$ , selling 11 units indeed maximizes our revenue!

STEP 14

Selling 11 units will maximize the weekly revenue.

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QuestionThe weekly revenue from the sale of $x$ units of a service is given by $R(x)=110 x-5 x^{2}$ thousand dollars, where $0 \leq x \leq 19$ . How many units should be sold to maximize the revenue?

Studdy Solution

STEP 1

What is this asking? We're trying to find the sweet spot of how many units of a service we need to sell to make the most money each week. Watch out! Don't mix up revenue with profit!
Revenue is just the total money coming in, not accounting for costs.

STEP 2

1. Define the function
2. Find the derivative
3. Set the derivative to zero
4. Solve for $x$
5. Check the boundaries

STEP 3

Our revenue function, $R(x)$ , tells us how much money we make each week based on the number of units, $x$ , we sell.
It's given by $R(x) = 110x - 5x^2$ , where $R(x)$ is in thousands of dollars.
Remember, $x$ has to be between 0 and 19 inclusive!

STEP 4

To find where our revenue is maximized, we need to find the derivative of our revenue function, $R'(x)$ .
The derivative tells us how much the revenue is changing for each additional unit sold.

STEP 5

Using the power rule, the derivative of $110x$ is $110$ , and the derivative of $-5x^2$ is $-10x$ .
So, our derivative is $R'(x) = 110 - 10x$ .

STEP 6

When the derivative is zero, it means the revenue isn't increasing or decreasing – it's at a peak (or a valley)!
We want the peak, the maximum revenue.
So, we set $R'(x) = 0$ .
That gives us $110 - 10x = 0$ .

STEP 7

Now, let's solve for $x$ !
We can add $10x$ to both sides of the equation: $110 = 10x$ .

STEP 8

Then, divide both sides by 10 to get $x = 11$ .
This tells us that selling 11 units might give us the maximum revenue.

STEP 9

Remember how $x$ has to be between 0 and 19?
We need to check the revenue at these boundaries, just in case they give us higher revenue than selling 11 units.

STEP 10

If $x = 0$ , $R(0) = 110 \cdot 0 - 5 \cdot 0^2 = 0$ .
Well, selling zero units gives us zero revenue, which makes sense!

STEP 11

If $x = 19$ , $R(19) = 110 \cdot 19 - 5 \cdot 19^2 = 2090 - 1805 = 285$ .
That's $\$285,000$ !

STEP 12

Now, let's check our potential maximum at $x = 11$ : $R(11) = 110 \cdot 11 - 5 \cdot 11^2 = 1210 - 605 = 605$ .
That's $\$605,000$ !

STEP 13

Since $\$605,000$ is greater than both $\$0$ and $\$285,000$ , selling 11 units indeed maximizes our revenue!

STEP 14

Selling 11 units will maximize the weekly revenue.

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QuestionThe weekly revenue from the sale of xxx units of a service is given by R(x)=110x−5x2R(x)=110 x-5 x^{2}R(x)=110x−5x2 thousand dollars, where 0≤x≤190 \leq x \leq 190≤x≤19. How many units should be sold to maximize the revenue?

Studdy Solution

STEP 1

What is this asking? We're trying to find the sweet spot of how many units of a service we need to sell to make the most money each week. Watch out! Don't mix up revenue with profit!Revenue is just the total money coming in, not accounting for costs.

STEP 2

1. Define the function2. Find the derivative3. Set the derivative to zero4. Solve for xxx5. Check the boundaries

STEP 3

Our revenue function, R(x)R(x)R(x), tells us how much money we make each week based on the number of units, xxx, we sell.It's given by R(x)=110x−5x2R(x) = 110x - 5x^2R(x)=110x−5x2, where R(x)R(x)R(x) is in thousands of dollars.Remember, xxx has to be between **0** and **19** inclusive!

STEP 4

To find where our revenue is maximized, we need to find the *derivative* of our revenue function, R′(x)R'(x)R′(x).The derivative tells us how much the revenue is changing for each additional unit sold.

STEP 5

Using the power rule, the derivative of 110x110x110x is 110110110, and the derivative of −5x2-5x^2−5x2 is −10x-10x−10x.So, our derivative is R′(x)=110−10xR'(x) = 110 - 10xR′(x)=110−10x.

STEP 6

When the derivative is zero, it means the revenue isn't increasing or decreasing – it's at a peak (or a valley)!We want the peak, the maximum revenue.So, we set R′(x)=0R'(x) = 0R′(x)=0.That gives us 110−10x=0110 - 10x = 0110−10x=0.

STEP 7

Now, let's solve for xxx!We can add 10x10x10x to both sides of the equation: 110=10x110 = 10x110=10x.

STEP 8

Then, divide both sides by **10** to get x=11x = 11x=11.This tells us that selling **11** units might give us the maximum revenue.

STEP 9

Remember how xxx has to be between **0** and **19**?We need to check the revenue at these boundaries, just in case they give us higher revenue than selling **11** units.

STEP 10

If x=0x = 0x=0, R(0)=110⋅0−5⋅02=0R(0) = 110 \cdot 0 - 5 \cdot 0^2 = 0R(0)=110⋅0−5⋅02=0.Well, selling zero units gives us zero revenue, which makes sense!

STEP 11

If x=19x = 19x=19, R(19)=110⋅19−5⋅192=2090−1805=285R(19) = 110 \cdot 19 - 5 \cdot 19^2 = 2090 - 1805 = 285R(19)=110⋅19−5⋅192=2090−1805=285.That's $285,000\$285,000$285,000!

STEP 12

Now, let's check our potential maximum at x=11x = 11x=11: R(11)=110⋅11−5⋅112=1210−605=605R(11) = 110 \cdot 11 - 5 \cdot 11^2 = 1210 - 605 = 605R(11)=110⋅11−5⋅112=1210−605=605.That's $605,000\$605,000$605,000!

STEP 13

Since $605,000\$605,000$605,000 is greater than both $0\$0$0 and $285,000\$285,000$285,000, selling **11** units indeed maximizes our revenue!

STEP 14

Selling **11** units will maximize the weekly revenue.

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Start learning now

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

QuestionThe weekly revenue from the sale of $x$ units of a service is given by $R(x)=110 x-5 x^{2}$ thousand dollars, where $0 \leq x \leq 19$ . How many units should be sold to maximize the revenue?

What is this asking? We're trying to find the sweet spot of how many units of a service we need to sell to make the most money each week. Watch out! Don't mix up revenue with profit!
Revenue is just the total money coming in, not accounting for costs.

1. Define the function
2. Find the derivative
3. Set the derivative to zero
4. Solve for $x$
5. Check the boundaries

Our revenue function, $R(x)$ , tells us how much money we make each week based on the number of units, $x$ , we sell.
It's given by $R(x) = 110x - 5x^2$ , where $R(x)$ is in thousands of dollars.
Remember, $x$ has to be between 0 and 19 inclusive!

To find where our revenue is maximized, we need to find the derivative of our revenue function, $R'(x)$ .
The derivative tells us how much the revenue is changing for each additional unit sold.

Using the power rule, the derivative of $110x$ is $110$ , and the derivative of $-5x^2$ is $-10x$ .
So, our derivative is $R'(x) = 110 - 10x$ .

When the derivative is zero, it means the revenue isn't increasing or decreasing – it's at a peak (or a valley)!
We want the peak, the maximum revenue.
So, we set $R'(x) = 0$ .
That gives us $110 - 10x = 0$ .

Now, let's solve for $x$ !
We can add $10x$ to both sides of the equation: $110 = 10x$ .

Then, divide both sides by 10 to get $x = 11$ .
This tells us that selling 11 units might give us the maximum revenue.

Remember how $x$ has to be between 0 and 19?
We need to check the revenue at these boundaries, just in case they give us higher revenue than selling 11 units.

If $x = 0$ , $R(0) = 110 \cdot 0 - 5 \cdot 0^2 = 0$ .
Well, selling zero units gives us zero revenue, which makes sense!

If $x = 19$ , $R(19) = 110 \cdot 19 - 5 \cdot 19^2 = 2090 - 1805 = 285$ .
That's $\$285,000$ !

Now, let's check our potential maximum at $x = 11$ : $R(11) = 110 \cdot 11 - 5 \cdot 11^2 = 1210 - 605 = 605$ .
That's $\$605,000$ !

Since $\$605,000$ is greater than both $\$0$ and $\$285,000$ , selling 11 units indeed maximizes our revenue!

Selling 11 units will maximize the weekly revenue.