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Question: If the total revenue for a hammer is \( R = 36x - .01{x^2} \) , then sale of how many hammers, \( x ...

If the total revenue for a hammer is R=36x−.01x2R = 36x - .01{x^2} , then sale of how many hammers, xx , will maximize the total revenue in dollars?

Explanation

Solution

Hint : We are given a function for the total revenue for hammers. The given function is a quadratic expression. The variable xx represents the number of hammers sold. Since the function is quadratic, the revenue will have a maximum or minimum value for some value of xx . We can find the maximum or minimum by finding the derivative of the function.

Complete step by step solution:
We have been given a function for the total revenue for a hammer given as R=36x−.01x2R = 36x - .01{x^2} .
We have to find the number of hammers sold for which the total revenue will be maximum.
Since the function of the total revenue is quadratic, this function will have a maximum or minimum value. The maximum or minimum of a function exists at the point where the derivative is zero.
The derivative of RR can be found as,
R′=d(36x−.01x2)dx=36−0.02xR' = \dfrac{{d\left( {36x - .01{x^2}} \right)}}{{dx}} = 36 - 0.02x
The value of xx at which the function is maximum or minimum is,
R′=0 ⇒36−0.02x=0 ⇒0.02x=36 ⇒x=360.02=1800   R' = 0 \\\ \Rightarrow 36 - 0.02x = 0 \\\ \Rightarrow 0.02x = 36 \\\ \Rightarrow x = \dfrac{{36}}{{0.02}} = 1800 \;
We also have to check the second derivative of the function at this point to determine whether the function attains maximum or minimum value at this point. The maximum second derivative should be negative.
R′′=d(36−0.02x)dx=−0.02R'' = \dfrac{{d\left( {36 - 0.02x} \right)}}{{dx}} = - 0.02
Since the first derivative is zero at x=1800x = 1800 and the second derivative is negative, the function attains maximum value at x=1800x = 1800 .
Hence, 18001800 hammers are to be sold to maximize the revenue.
So, the correct answer is “ x=1800x = 1800 ”.

Note : We found the maximum of the function at the point where the first derivative is zero. We also have to check the second derivative of the function at that point to determine whether it is a maxima or minima point. We can check by graph of the function that the revenue increases upto x=1800x = 1800 and after that revenue starts decreasing, thus attaining maximum value at this point.