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The price p (in dollars) and the quantity x sold of a certain product obey the demand equation p= -1/6x + 100. Find a model that expresses the revenue R as a function of x.

The Price P In Dollars And The Quantity X Sold Of A Certain Product Obey The Demand Equation

The main objective of this question is to find the revenue model of the given equation as just a function with respect to x.

This question uses the concept of revenue model. A revenue model is a blueprint that outlines how a startup company will generate revenue or annual profit out of its basic business operations. Revenue is a blueprint that outlines how a startup business would then generate revenue or annual profit out of its standard daily operations, as well as how it will cover operating costs and expenses.

Expert Answer

We have to find the revenue model for the given expression. A revenue model is a blueprint that outlines how a startup company will generate revenue or annual profit out of its basic business operations. The given expression is:

\[p \space = \space – \space \frac{1}{6}x \space  + \space 100 \]

We know that:

\[R \space = \space xp \]

So:

\[R \space = \space x (- \space \frac{1}{6}x \space + \space 100 ) \]

Multiplying $ x $ results in:

\[R \space = \space – \space \frac{1}{6}x^2 \space + \space 100 x \]

Hence, the final answer is:

\[R \space = \space – \space \frac{1}{6}x^2 \space + \space 100 x \]

Numerical Answer

The revenue model for the given expression $ p =  –  \frac{1}{6}x  +  100 $ where p is the price in dollars and the quantity of product sold is $ x  $ :

\[R \space = \space – \space \frac{1}{6}x^2 \space + \space 100 x \]

Example

Find the revenue model for the two  expressions $ p =  –  \frac{1}{8}x  +  120 $ and  $ p =  –  \frac{1}{8}x ^2 +  220 $ \space where $ p $ is the price in dollars and the quantity of product sold is $ x  $ .

We have to find the revenue model for the given expression which is:

\[p \space = \space – \space \frac{1}{8}x \space  + \space 120 \]

where $ p $ is the price in dollars and the quantity of product sold is $ x  $.

We know that:

\[R \space = \space xp \]

So:

\[R \space = \space x (- \space \frac{1}{8}x \space + \space 120 ) \]

Multiplying $ x $ results in:

\[R \space = \space – \space \frac{1}{8}x^2 \space + \space 120 x \]

Hence, the final answer is:

\[R \space = \space – \space \frac{1}{8}x^2 \space + \space 120 x \]

Now for the second expression which is:

\[p \space = \space – \space \frac{1}{8}x ^2 +  220 \]

where $ p $ is the price in dollars and the quantity of product sold is $ x  $

We have to find the revenue model for the given expression, which is:

\[p \space = \space – \space \frac{1}{8}x^2 \space  + \space 220 \]

We know that:

\[R \space = \space xp \]

So:

\[R \space = \space x (- \space \frac{1}{8}x^2 \space + \space 220 ) \]

Multiplying $ x $ results in:

\[R \space = \space – \space \frac{1}{8}x^3 \space + \space 220 x \]

Thus, the final answer is:

\[R \space = \space – \space \frac{1}{8}x^3 \space + \space 220 x \]

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