Find the quantity that produces maximum profit, given demand function 
and cost function 
| Foundations:
|
Recall that the demand function,  , relates the price per unit  to the number of units sold,  .
Moreover, we have several important important functions:
|
 , the total cost to produce units;
 , the total revenue (or gross receipts) from producing units;
 , the total profit from producing units.
|
| In particular, we have the relations
|
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P(x)=R(x)-C(x),}
|
| and
|
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R(x)=x\cdot p(x).}
|
| Using these equations, we can find the maximizing production level by determining when the first derivative of profit is zero.
|
Solution:
| Step 1:
|
| Find the Profit Function: We have
|
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R(x)\,=\,x\cdot p(x)\,=\,x\cdot (70-3x)\,=\,70x-3x^2.}
|
| From this,
|
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P(x)\,=\,R(x)-C(x)\,=\,70x-3x^2- \left(120 - 30x + 2x^2 \right)\,=\,-120 + 100 x - 5 x^2 .}
|
| Step 2:
|
| Find the Maximum: The equation for marginal revenue is
|
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P(x)\,=\,-120 + 100 x - 5 x^2 .}
|
| Applying our power rule to each term, we find
|
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P'(x)\,=\,100-10x\,=\,10(10-x).}
|
| The only root of this occurs at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=10}
, and this is our production level to achieve maximum profit.
|
|
|
| Final Answer:
|
| Maximum profit occurs when we produce 10 items.
|
Return to Sample Exam