MATEMATİK I

200+2x²+4x = 350-x²+19x

3x²-15x-150=0

and dividing both sides by 3 we get:

x²-5x-50=0 so:

(x-10)(x+5)=0

Therefore x=10 and x=-5 are the roots of this equation. Since production amount can not be negative, the answer is x=10.

MC=dTC/dx=6x+5

So:

6x+5=95

6x=90

x=15

Thus profit function P(x)=TR(x)-TC(x)=2x²+x - (100+x²+17x)=x²-16x-100

For maximum profit first derivative must be equal to zero. Thus:

dP(x)/dx=2x-16=0

Thus x=8

Since marginal revenue is the first derivative of the total revenue:

MR=dTR/dx=100-4x

and for x=6, MR=100-4*6=76

so we first have to write x as a function of p.

Therefore:

p=100-2x

2x=100-p

x=50-0.5p

Thus dx/dp=-0.5

And for p=40, x=50-0.5*40=30

Thus for x=30 and p=40;

price elasticity of demand=(dx/dp)*(p/x)=-0.5*40/30=-20/30=-2/3

The labor cost will be maximum when the first derivative is equal to zero. Thus:

dLC/dw=300-4w=0

w=75

MC=dTC/dx=3x²-6x

For MC be decreasing, its first derivative, namely dMC/dx must be negative.

Thus:

dMC/dx=6x-6

So for 6x-6<0

6x<6

x<1

Thus 5+4p=35-6p

10p=30

p=3

S(p)=30+20p = D(p)=150-10p

120=30p

4=p

Since only 3 is less than 4 among the answers, excess demand occurs only when price is equal to 3.