Question

Some years ago it was estimated that the demand for steel approximately satisfied the equation p equals 201 minus 20 x​, and the total cost of producing x units of steel was Upper C (x )equals 153 plus 41 x. ​(The quantity x was measured in millions of tons and the price and total cost were measured in millions of​ dollars.) Determine the level of production and the corresponding price that maximize the profits.

Answer 1

Answer: The level of production is 4 units and the corresponding price that maximize the profits is $121.

Step-by-step explanation:

Since we have given that

the demand for steel would be

$p=201-20x$

$R(x)=xp(x)=x(201-20x)=201x-20x^2$

Total cost of producing x units of steel would be

$C(x)=153+41x$

Profit = R(x)-C(x)

So, it becomes,

$profit=201x-20x^2-153-41x\\\\P(x)=-20x^2+160x-153$

First we derivative w.r.t to x we get that

$P'(x)=-40x+160$

Now, we will find the critical points,

$-40x+160=0\\\\-40x=-160\\\\x=\dfrac{-160}{-40}\\\\x=4$

So, now second derivative w.r.t. x.

We get that

$P''(x)=-40$

So, there will be maximum profit at 4 units of level of production  and the corresponding price that maximize the profit would be

$p(x)=201-20x\\\\p(4)=201-20(4)\\\\p(4)=201-80\\\\p(4)=121$

Hence, the level of production is 4 units and the corresponding price that maximize the profits is $121.