Assignment 7: Maximum of a profit function.

Assignment 7: Maximum of a profit function.

Based on the assignment, we got the:
TR=1400Q -7.5Q^2
TC=Q^3-6Q^2+140Q+750

Revenue= TR-TC
So, we can get the function of profit is :
Profit=1260Q-1.5Q^2-Q^3-750

MR=1400-15Q
MC=3Q^2-12Q+140

We can draw the graph of TR,TC,PROFIT ,MR and MC.

When MC=MR , the company achieve the most profit.

According to the graph of MC and MR, we can see that when Q=20, MC=MR.
When Q=20 the profit is biggest. Profit=15850

Assignmet 6: Tipping (braking) point of a product life cycle.

Assignmet 6: Tipping (braking) point of a product life cycle.

TC=Q^3-12Q^2+60Q

According to the TC function, we can get the MC and AC function.
MC=3Q^2-24Q+60
AC=Q^2-12Q+60

When we have those product function, we can draw the graph.

When MC=AC, we can get Q=6 and MC=AC=24, then when Q=6 TC=144.
If Q<6, the TC graph is degressive increasing.
If Q>6, the TC graph is progressive increasing.
So the Point (6,144) is the tipping point of  TC.
We can get the tangent could assume be y=24x.

y=24+25 is the point (1.49) tangent of TC. It paralla with y=24x.


After point(6.144). the MC>AC, the productor could consider keep on product, because the cost is progressive increasing.