Given information:

The given information is

Commodity is C(x)=1800+25x−0.2x2+0.001x3

Demand function is p(x)=48.2−0.03x .

Calculation:

The cost function is C(x)=1800+25x−0.2x2+0.001x3

Demand function is p(x)=48.2−0.03x

The revenue function R(x)=x.p(x)

Or R(x)=48.2x−0.03x2

Now we draw the curve of C(x) and R(x) on the same screen and get that Manufacturers has profit when R>C , and maximum profit occurs when R and C have parallel tangents.

That is at the production level of x=160 units.

Therefore, the production level of x=160 units will be maximum profit.

Given information:

The given information is

Commodity is C(x)=1800+25x−0.2x2+0.001x3

Demand function is p(x)=48.2−0.03x .

Calculation:

Now the marginal revenue function is

  R'(x)=48.2−0.06x

And marginal cost function is

  C'(x)=25x−0.4x+0.003x3

For maximize the profit we must have

  R'(x)=C'(x)

Or 48.2−0.06x=25x−0.4x+0.003x30.003x2−0.34x−23.2=0

The root of the equation ax2+bx+c=0

  x=−b±b2−4ac2a

Here we are not considering x=−b−b2−4ac2a because x>0 .

Thus we have

  x=0.34+(−0.34)2−4×0.003×23.20.006

  x=161 Units.

Since R'(x)= −0.06 and C'(x)= 0.006x−0.4

So R'(x)<C(x) for x=161 units

Therefore, the production level 161 units will maximize the profit.

Given information:

The given information is

Commodity is C(x)=1800+25x−0.2x2+0.001x3

Demand function is p(x)=48.2−0.03x .

Calculation:

Now the average cost function is

  c(x)=C(x)x=1800x+25−0.2x+0.001x2

And marginal cost is

  C'(x)=25x−0.4x+0.003x3

For minimum average cost we must have

  c(x)=C'(x)

  1800x+25−0.2x+0.001x2=25x−0.4x+0.003x30.002x3−0.2x2=1800

By graphical method

We draw the curve f(x)=0.002x3−0.2x2 and y=1800 .

We estimate the point of intersection of these curves. The x-coordinate of the point of intersection is about 144 units.

So at the production level of 144 units the average cost is minimum.

Now,

  c(x)=C(x)x=1800x+25−0.2x+0.001x2

Then c'(x)=1800x2−0.2+0.002x

And c''(x)=2×1800x3+0.002 >0 for all x>0

So at x=144, c(x) has an absolute minimum.