Question

Find the number of units that produce a maximum revenue, R =95x-0.1x^2, where R is the total revenue in dollars and x is the number of units sold

Answer 1

Answer:

$x=475$ units

Step-by-step explanation:

Method 1

The given revenue function is $R(x)=95x-0.1x^2$.

We rewrite this function in the vertex form by completing the square.

$\Rightarrow R(x)=-0.1x^2+95x$.

$\Rightarrow R(x)=-0.1(x^2-950x)+0$.

We add and subtract half the coefficient of $x$ multiplied by a factor of $-0.1$, which is $-0.1(-\frac{950}{2})^2=-0.1( -475)^2$ to get,

$R(x)=-0.1(x^2-950x)+-0.1(-475)^2--0.1(-475)^2 +0$.

We factor $-0.1$ out of the first two expressions again to get,

$R(x)=-0.1(x^2-950x+(-475)^2)--0.1(-475)^2 +0$.

We now got a perfect square.

$\Rightarrow R(x)=-0.1(x-475)^2+22562.5$.

Therefore the maximum revenue occurs when $x=475$ units were sold.

Method 2

Use derivatives to find the x-value of the maximum point.

$R(x)=-0.1x^2+95x$

$R'(x)=-0.2x+95$

At maximum point,

$R'(x)=0$

$\Rightarrow -0.2x+95=0$.

$\Rightarrow -0.2x=-95$.

$\Rightarrow x=\frac{-95}{-0.2}$.

$\Rightarrow x=475$.