QuestionFind the price $P$ that maximizes revenue $TR$ given the demand function $Q=90-1.5P$ .

Studdy Solution

STEP 1

Assumptions1. The demand function is given by $Q=90-1.5$ . and $Q$ represent the price and quantity, respectively3. Revenue $TR$ is given by the product of price and quantity $Q$ , i.e., $TR=PQ$

STEP 2

First, we need to express the revenue $TR$ in terms of price $$. We can do this by substituting the demand function into the revenue function.
$TR =(90-1.5)$

STEP 3

Expand the right side of the equation.
$TR =90 -1.5^2$

STEP 4

To find the maximum revenue, we need to find the derivative of the revenue function with respect to price $$ and set it equal to zero. This is because the maximum or minimum of a function occurs where its derivative is zero.
$\frac{dTR}{d} =90 -3$

STEP 5

Set the derivative equal to zero and solve for $$.
$90 -3 =0$

STEP 6

olve the equation for $$.
$= \frac{90}{3} =30$ At a price of $30, the revenue$ TR$ is at maximum.

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QuestionFind the price $P$ that maximizes revenue $TR$ given the demand function $Q=90-1.5P$ .

Studdy Solution

STEP 1

Assumptions1. The demand function is given by $Q=90-1.5$ . and $Q$ represent the price and quantity, respectively3. Revenue $TR$ is given by the product of price and quantity $Q$ , i.e., $TR=PQ$

STEP 2

First, we need to express the revenue $TR$ in terms of price $$. We can do this by substituting the demand function into the revenue function.
$TR =(90-1.5)$

STEP 3

Expand the right side of the equation.
$TR =90 -1.5^2$

STEP 4

To find the maximum revenue, we need to find the derivative of the revenue function with respect to price $$ and set it equal to zero. This is because the maximum or minimum of a function occurs where its derivative is zero.
$\frac{dTR}{d} =90 -3$

STEP 5

Set the derivative equal to zero and solve for $$.
$90 -3 =0$

STEP 6

olve the equation for $$.
$= \frac{90}{3} =30$ At a price of $30, the revenue$ TR$ is at maximum.

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QuestionFind the price PPP that maximizes revenue TRTRTR given the demand function Q=90−1.5PQ=90-1.5PQ=90−1.5P.

Studdy Solution

STEP 1

Assumptions1. The demand function is given by Q=90−1.5Q=90-1.5Q=90−1.5 . and $Q$ represent the price and quantity, respectively3. Revenue $TR$ is given by the product of price and quantity QQQ, i.e., TR=PQTR=PQTR=PQ

STEP 2

First, we need to express the revenue TRTRTR in terms of price $$. We can do this by substituting the demand function into the revenue function.TR=(90−1.5)TR =(90-1.5)TR=(90−1.5)

STEP 3

Expand the right side of the equation.TR=90−1.52TR =90 -1.5^2TR=90−1.52

STEP 4

To find the maximum revenue, we need to find the derivative of the revenue function with respect to price $$ and set it equal to zero. This is because the maximum or minimum of a function occurs where its derivative is zero.dTRd=90−3\frac{dTR}{d} =90 -3ddTR​=90−3

STEP 5

Set the derivative equal to zero and solve for $$.90−3=090 -3 =090−3=0

STEP 6

olve the equation for $$.=903=30 = \frac{90}{3} =30=390​=30At a price of 30,therevenue30, the revenue 30,therevenueTR$ is at maximum.

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

QuestionFind the price $P$ that maximizes revenue $TR$ given the demand function $Q=90-1.5P$ .

Assumptions1. The demand function is given by $Q=90-1.5$ . and $Q$ represent the price and quantity, respectively3. Revenue $TR$ is given by the product of price and quantity $Q$ , i.e., $TR=PQ$

First, we need to express the revenue $TR$ in terms of price $$. We can do this by substituting the demand function into the revenue function.
$TR =(90-1.5)$

Expand the right side of the equation.
$TR =90 -1.5^2$

To find the maximum revenue, we need to find the derivative of the revenue function with respect to price $$ and set it equal to zero. This is because the maximum or minimum of a function occurs where its derivative is zero.
$\frac{dTR}{d} =90 -3$

Set the derivative equal to zero and solve for $$.
$90 -3 =0$

olve the equation for $$.
$= \frac{90}{3} =30$ At a price of $30, the revenue$ TR$ is at maximum.