
Find the maximum profit that a company can make if the profit function is given by:
\[P\left( x \right)=41-72x-18{{x}^{2}}\]
Answer
600.3k+ views
Hint: We will use the derivative method to find the maximum value of a function. When a function’s slope is zero at x and the second derivative at x is:
(a) less than zero, it is local maximum
(b) greater than zero, it is local minimum
(c) equal to zero, can’t say
We know that the slope of a function F(x) is equal to its derivative i.e. F’(x).
Complete step-by-step answer:
We have been asked to find the profit of a company if the profit function is given by:
\[P\left( x \right)=41-72x-18{{x}^{2}}\]
We will use the derivative method to find the maximum value of P(x).
When a function’s slope is zero at x and second derivative at x is:
less than zero, then it is the local maximum.
We know that the slope of a function f(x) is equal to the derivative of the function, i.e. f’(x).
\[\begin{align}
& P'\left( x \right)=\dfrac{d}{dx}\left( 41-72x-18{{x}^{2}} \right)=\dfrac{d}{dx}\left( 41 \right)-\dfrac{d}{dx}\left( 72x \right)-\dfrac{d}{dx}\left( 18{{x}^{2}} \right) \\
& \Rightarrow P'(x)=0-72-36x \\
\end{align}\]
Now, \[P'(x)=0\]
\[\begin{align}
& \Rightarrow -72-36x=0 \\
& \Rightarrow -72=36x \\
& \Rightarrow \dfrac{-72}{36}=x \\
& \Rightarrow -2=x \\
\end{align}\]
Now the second derivative of P(x) is given as follows:
\[P''(x)=\dfrac{d}{dx}\left( P'(x) \right)=\dfrac{d}{dx}\left( -71-36x \right)=\dfrac{d}{dx}\left( -72 \right)-\dfrac{d}{dx}\left( 36x \right)=0-36<0\]
Hence we get the second derivative of P(x)is always negative.
\[\Rightarrow P''(-2)\] is also a negative value.
\[\Rightarrow \]At \[x = -2\] there is a local maxima of the function which means we get the maximum value at x= -2.
Maximum value at x= -2 is given by:
\[P(-2)=41-72\times -2-18\times {{(-2)}^{2}}=41+144-72=113\]
Therefore, the maximum profit that a company can make is 113.
Note: Be careful while finding the first and the second derivative of the function as there is a chance of calculation mistake during differentiation. Also, remember that the first derivative of any function is also known as slope of the function. Also, remember that the value of ‘x’ that is obtained by equating P’(x)=0, is known as a critical point.
(a) less than zero, it is local maximum
(b) greater than zero, it is local minimum
(c) equal to zero, can’t say
We know that the slope of a function F(x) is equal to its derivative i.e. F’(x).
Complete step-by-step answer:
We have been asked to find the profit of a company if the profit function is given by:
\[P\left( x \right)=41-72x-18{{x}^{2}}\]
We will use the derivative method to find the maximum value of P(x).
When a function’s slope is zero at x and second derivative at x is:
less than zero, then it is the local maximum.
We know that the slope of a function f(x) is equal to the derivative of the function, i.e. f’(x).
\[\begin{align}
& P'\left( x \right)=\dfrac{d}{dx}\left( 41-72x-18{{x}^{2}} \right)=\dfrac{d}{dx}\left( 41 \right)-\dfrac{d}{dx}\left( 72x \right)-\dfrac{d}{dx}\left( 18{{x}^{2}} \right) \\
& \Rightarrow P'(x)=0-72-36x \\
\end{align}\]
Now, \[P'(x)=0\]
\[\begin{align}
& \Rightarrow -72-36x=0 \\
& \Rightarrow -72=36x \\
& \Rightarrow \dfrac{-72}{36}=x \\
& \Rightarrow -2=x \\
\end{align}\]
Now the second derivative of P(x) is given as follows:
\[P''(x)=\dfrac{d}{dx}\left( P'(x) \right)=\dfrac{d}{dx}\left( -71-36x \right)=\dfrac{d}{dx}\left( -72 \right)-\dfrac{d}{dx}\left( 36x \right)=0-36<0\]
Hence we get the second derivative of P(x)is always negative.
\[\Rightarrow P''(-2)\] is also a negative value.
\[\Rightarrow \]At \[x = -2\] there is a local maxima of the function which means we get the maximum value at x= -2.
Maximum value at x= -2 is given by:
\[P(-2)=41-72\times -2-18\times {{(-2)}^{2}}=41+144-72=113\]
Therefore, the maximum profit that a company can make is 113.
Note: Be careful while finding the first and the second derivative of the function as there is a chance of calculation mistake during differentiation. Also, remember that the first derivative of any function is also known as slope of the function. Also, remember that the value of ‘x’ that is obtained by equating P’(x)=0, is known as a critical point.
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