C(x)=5,000+20x-1/4x^2 is a cost function. Calculate the marginal

Trent Carpenter 2021-10-31 Answered
\(\displaystyle{C}{\left({x}\right)}={5},{000}+{20}{x}-\frac{{1}}{{4}}{x}^{{2}}\)
is a cost function.
Calculate the marginal average cost when 20 (a), use your result to estimate the total cost when x=21 (b) and calculate the percentage error between this estimate and the actual cost when x=21 (c).

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Expert Answer

lamusesamuset
Answered 2021-11-01 Author has 17890 answers
\(\displaystyle{C}{\left({x}\right)}={5},{000}+{20}{x}-\frac{{1}}{{4}}{x}^{{2}}\)
\(\displaystyle\frac{{{C}{\left({x}\right)}}}{{x}}=?\)
\(\displaystyle\frac{{{C}{\left({x}\right)}}}{{x}}\)
\(\displaystyle=\frac{{{5},{000}+{20}{x}-\frac{{1}}{{4}}{x}^{{2}}}}{{x}}\)
\(\displaystyle=\frac{{5000}}{{x}}+{20}-\frac{{1}}{{4}}{x}\)
\(\displaystyle\overline{{C}}'{\left({x}\right)}=\frac{{{d}{\left(\frac{{5000}}{{x}}+{20}-\frac{{1}}{{4}}{x}\right)}}}{{{\left.{d}{x}\right.}}}\)
\(\displaystyle=-\frac{{5000}}{{x}^{{2}}}-\frac{{1}}{{4}}\)
So \(\displaystyle\overline{{C}}'{\left({20}\right)}=-\frac{{5000}}{{x}^{{2}}}-\frac{{1}}{{4}}\)
\(\displaystyle=-{12.5}-{0.25}\)
\(\displaystyle=-{12.75}\)
And \(\displaystyle{C}{\left({20}\right)}=\frac{{5000}}{{20}}+{20}-\frac{{1}}{{4}}{\left({20}\right)}={265}\) (a)
\(\displaystyle\overline{{C}}{\left({21}\right)}=\overline{{C}}{\left({20}\right)}+\overline{{C}}'{\left({20}\right)}\)
\(\displaystyle={265}-{12.75}\)
\(\displaystyle={252.25}\)
\(\displaystyle{C}\frac{{{21}}}{{21}}={252.25}\)
\(\displaystyle{C}{\left({21}\right)}={5297.25}\) (b)
\(\displaystyle{\underset{{{E}}}{{{C}}}}={5000}+{20}{\left({21}\right)}-\frac{{1}}{{4}}{\left({21}\right)}^{{2}}\)
\(\displaystyle={5309.75}\)
\(\displaystyle\frac{{{5309.75}-{5297.25}}}{{5309.75}}\times{100}={0.24}\%\) (c)
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(b) and calculate the percentage error between this estimate and the actual cost when \(x=21 \)

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