Find the quantity that produces maximum profit, given the demand function 
and cost function 
.
| Foundations:
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Recall that the demand function,  , relates the price per unit  to the number of units sold,  .
Moreover, we have several important important functions:
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 , the total cost to produce 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}
units;
- 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)}
, the total revenue (or gross receipts) from producing 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}
units;
- 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)}
, the total profit from producing 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}
units.
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| In particular, we have the relations
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- 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),}
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| and
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- 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).}
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| Using these equations, we can find the maximizing production level by determining when the first derivative of profit is zero.
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Solution:
| Step 1:
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| Find the Profit Function: We have
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- 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 (90-3x)\,=\,90x-3x^2.}
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| From this,
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- 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)\,=\,90x-3x^2- \left(200-30x+x^2 \right)\,=\,120x-4x^2-200 .}
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| Step 2:
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| Find the Maximum: The equation for marginal revenue is
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- 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)\,=\,120x-4x^2-200 .}
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| Applying our power rule to each term, we find
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- 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)\,=\,120-8x\,=\,8(15-x).}
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| 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=15}
, and this is our production level to achieve maximum profit.
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|
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| Final Answer:
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| Maximum profit occurs when we produce 15 items.
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