Question

How do you find the instantaneous rate of change of revenue when 1000 units are produced if the revenue (in thousands of dollars) from producing $x$ units of an item is $R\left( x \right) = 12x - 0.005{x^2}$?

Answer 1

In this question we have to find the instantaneous rate of change of revenue, we will find the instantaneous rate of change of a function at a point by finding the derivative of that function and plugging in the $x$-value of the point. Here we will derive the given function using derivative formulas, and then by substituting the given value i.e, 1000 in the differentiated function we will get the required rate of change.

Complete step by step solution:
Given that the revenue from producing $x$ units of an item is $R\left( x \right) = 12x - 0.005{x^2}$, and the units produced are 1000,
Now derive the given function, we get,
$\Rightarrow \dfrac{d}{{dx}}R\left( x \right) = \dfrac{d}{{dx}}12x - 0.005{x^2}$,
Now using distributive property we get,
$\Rightarrow R'\left( x \right) = \dfrac{d}{{dx}}12x - \dfrac{d}{{dx}}0.005{x^2}$,
Now taking out the constant terms we get,
$\Rightarrow R'\left( x \right) = 12\dfrac{d}{{dx}}x - 0.005\dfrac{d}{{dx}}{x^2}$,
Now deriving using derivative formulas, $\dfrac{d}{{dx}}x = 1$and $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$, we get,
$\Rightarrow R'\left( x \right) = 12\left( 1 \right) - 0.005\left( {2x} \right)$,
Now multiplying we get,
$\Rightarrow R'\left( x \right) = 12 - 0.01x$,
Now to find the rate of change when 1000 units are produced we have to substitute 1000 in place of $x$, we get
$\Rightarrow R'\left( {1000} \right) = 12 - 0.01\left( {1000} \right)$,
Now simplifying we get,
$\Rightarrow R'\left( {1000} \right) = 12 - 10$,
Now further simplification we get,
$\Rightarrow R'\left( {1000} \right) = 2$,

$\therefore$The instantaneous rate of change of revenue when 1000 units are produced will be equal to 2 thousand dollars per unit.

Note: Instantaneous rate of change of a function is represented by the slope of the line, it tells you by how much the function is increasing or decreasing as the $x$-values change. The instantaneous rate of change at a point is equal to the function's derivative evaluated at that point.