Question

How to draw the graph?

1. MPL=a+2bL-3cL^2
APL=a+bL-cL^2

2. MR=5-2Q
AR=5-Q

3.MC=6Q+7
AC=3Q+7+12/Q

I want detailed explanation??
thank you

Answer 1

Lets start with 2.

MR = 5 - 2Q

	5 - 2Q
Q	MR
1	3
2	1
3	-1

AR = 5 - Q

	5 - Q
Q	AR
1	4
2	3
3	2
4	1
5	0

3.

MC = 6Q + 7

	6Q + 7
Q	MC
1	13
2	19
3	25
4	31
5	37

AC = 3Q + 7 + 12/Q

	3Q + 7 + 12/Q
Q	AC
1	22
2	19
3	20
4	22
5	24.4

For 1st sub part

Lets find the output at which , MC = AC

6Q + 7 = 3Q + 7 + 12/Q

3Q = 12/Q

Q = 2

At Q=2, MC and AC both are 19.

Using the equations of MPL and APL now

Max value of APL will be when

differentiation of equation of APL = 0

d/dL(a+bL-cL^2) = 0

b - 2cL = 0

L = b/2c

At L = b/2c, we will get maximum value of APL.

At the same point APL and MPL will intersect as well.

Substituting L = b/2c in both equations of MPL and APL

APL = a + b(b/2c) - c(b/2c)^2 = a + b^2/2c - b^2/4c = a + b^2/4c

MPL = a + 2b(b/2c) - 3c(b/2c)^2 = a + b^2/c - 3b^2/4c = a + b^2/4c

In the above equation, it can be said that Max APL is at L2.

So, L2 = b/2c