
The demand function for a certain commodity is $p\left( x \right) = 10 - 0.001x$, where p is measured in dollars and x is the number of units produced and sold. The told cost of producing x items is $C\left( x \right) = 50 + 5x$ . Determine the level of production that maximizes the profit.
Answer
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Hint: We will use the definition of the profit function, which is, $P\left( x \right) = xp\left( x \right) - C\left( x \right)$ and substitute the given values in it. Then we will find the critical point by equating the first derivative of the profit function to 0. In the end, we just need to check whether the critical point obtained is the point of maximum or minimum depending on the sign of the second derivative of the profit function.
Complete step-by-step answer:
Let us consider the things given to us.
$p\left( x \right) = 10 - 0.001x$
$C\left( x \right) = 50 + 5x$
Our objective is to determine the level of production that maximizes the profit.
It is known that the function to determine the profit is given by;
$P\left( x \right) = xp\left( x \right) - C\left( x \right)$
Let us substitute the given values into the profit function and simplify it.
$
\Rightarrow P\left( x \right) = x\left( {10 - 0.001x} \right) - \left( {50 + 5x} \right) \\
\Rightarrow P\left( x \right) = 10x - 0.001{x^2} - 50 - 5x \\
\Rightarrow P\left( x \right) = 5x - 0.001{x^2} - 50 \\
$
Let us now differentiate the profit function.
$ \Rightarrow P'\left( x \right) = 5 - 0.002x$
Put this obtained expression for the derivative of the profit function to 0, to find the value of the critical point.
\[
\Rightarrow 5 - 0.002x = 0 \\
\Rightarrow x = \dfrac{5}{{0.002}} \\
\Rightarrow x = \dfrac{{5000}}{2} \\
\Rightarrow x = 2500 \\
\]
Thus, we get that 2500 is the critical point of the profit function.
Now we are just left to check whether the second derivative of the profit function is negative or not. If it is, then 2500 is the point of maximum.
Let us not calculate the second derivative of the profit function.
\[
\Rightarrow P''\left( x \right) = - 0.002 \\
\Rightarrow P''\left( x \right) < 0 \\
\]
Hence, we can conclude that 2500 is the point of maximum. Thus, the company has the largest profit when $x = 2500$.
Note: While finding the critical point for the profit function, put the first derivative of the function to be equal to zero and not the function itself. Also, once you have calculated the critical point, don’t forget to check if that is a point of maximum or minimum. This could be done by finding the second derivative of the profit function. Avoid making any calculation errors.
Complete step-by-step answer:
Let us consider the things given to us.
$p\left( x \right) = 10 - 0.001x$
$C\left( x \right) = 50 + 5x$
Our objective is to determine the level of production that maximizes the profit.
It is known that the function to determine the profit is given by;
$P\left( x \right) = xp\left( x \right) - C\left( x \right)$
Let us substitute the given values into the profit function and simplify it.
$
\Rightarrow P\left( x \right) = x\left( {10 - 0.001x} \right) - \left( {50 + 5x} \right) \\
\Rightarrow P\left( x \right) = 10x - 0.001{x^2} - 50 - 5x \\
\Rightarrow P\left( x \right) = 5x - 0.001{x^2} - 50 \\
$
Let us now differentiate the profit function.
$ \Rightarrow P'\left( x \right) = 5 - 0.002x$
Put this obtained expression for the derivative of the profit function to 0, to find the value of the critical point.
\[
\Rightarrow 5 - 0.002x = 0 \\
\Rightarrow x = \dfrac{5}{{0.002}} \\
\Rightarrow x = \dfrac{{5000}}{2} \\
\Rightarrow x = 2500 \\
\]
Thus, we get that 2500 is the critical point of the profit function.
Now we are just left to check whether the second derivative of the profit function is negative or not. If it is, then 2500 is the point of maximum.
Let us not calculate the second derivative of the profit function.
\[
\Rightarrow P''\left( x \right) = - 0.002 \\
\Rightarrow P''\left( x \right) < 0 \\
\]
Hence, we can conclude that 2500 is the point of maximum. Thus, the company has the largest profit when $x = 2500$.
Note: While finding the critical point for the profit function, put the first derivative of the function to be equal to zero and not the function itself. Also, once you have calculated the critical point, don’t forget to check if that is a point of maximum or minimum. This could be done by finding the second derivative of the profit function. Avoid making any calculation errors.
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