Question

The weekly revenue for a product is given by R(x)=307.8x−0.045x2, and the weekly cost is C(x)=10,000+153.9x−0.09x2+0.00003x3, where x is the number of units produced and sold.(a) How many units will give the maximum profit?(b) What is the maximum possible profit?

Answer 1

Answer:

The number of units that will give the maximum profit is;

$1900\text{ units}$

The maximum possible profit is;

$\text{ \ }239,090$

Explanation:

Given that the weekly revenue for a product is given by ;

$R(x)=307.8x-0.045x^2$

and the weekly cost is ;

$C(x)=10,000+153.9x-0.09x^2+0.00003x^3$

Recall that

Profit = Revenue - Cost

$P(x)=R(x)-C(x)$$\begin{gathered} P(x)=307.8x-0.045x^2-(10,000+153.9x-0.09x^2+0.00003x^3) \\ P(x)=307.8x-0.045x^2-10,000-153.9x+0.09x^2-0.00003x^3 \\ P(x)=153.9x+0.045x^2-0.00003x^3-10,000 \end{gathered}$

Using graph to derive the maximum point on the function;

Therefore, the maximum point is at the point;

$(1900,239090)$

So;

The number of units that will give the maximum profit is;

$1900\text{ units}$

The maximum possible profit is;

$\text{ \ }239,090$