E_p = |(p / q) (dq / dp)| = 1 for unit elastic ; dq / dp = -0.04 p ; Ep = |(p / (600 – (0.02 p²))) (-0.04 p)| = 1 ; p / (600 – (0.02 p2)) (-0.04 p) = 1 ; 0.04 p² / (600 – 0.02 p²) = 1 ; 0.06 p² = 600 ; p² = 10.000 ; p = 100 . pg. 166. Correct answer is C.

We first find the equilibrium point by equating the supply and demand functions. Therefore, the equality

gives 

Since there will be a shortage in the market, to the left of this point the correct answer is

Writing the demand function for x as x=25-1/4p, and writing the supply as x=2p+20 the equilibrium price may be found as 

25-1/4p=2p+20

which is written as

25-20=2p+1/4p

and the price is found as

p=20/9. Therefore, the answer is D.

If demand is of unit elasticity, i.e. |E_p|=1, revenue is unaffected by a small change in price.

If we denote the slope of the curve with m, then:

f(x)=e^3x

f'(x)=3e^3x

f(x)=9e^3x

We can find the shortest distance by drawing.

So the distance between points A and B is √2

We first find the equilibrium point by equating the supply and demand functions. Therefore, the equality 

-4x+150=6x-275, x=42,5. Because there will be a shortage in the market, to the left of this point the correct is x<42,5. Therefore, the answer is B.

velocity at time t=lim_h›0 (f(t+h)-f(t))/h

since f(t) = 2t² – t:

lim_h›0 (f(t+h)-f(t))/h=((2(t+h)²-(t+h))-(2t²-t))/h

=lim_h›0 (2t²+2h² +4th-t-h-2t²+t)/h

=lim_h›0 (2h² +4th-h)/h

=h(2h+4t-1)/h

=2h+4t-1

And for t=3 and h=0;

2h+4t-1=11

Since P(x)=R(x)-C(x), according to the given information we can write P(x)=(150-2x)x-(120+60x-x² )=-x²+90x-120. Now, since there is an extra of 4 TL tax the modified profit function is P(x)=-x²+(90-4)x-120=-x²+86x-120. The price that maximizes the profit is where the derivative is zero, i.e., P' (x)=-2x+86=0 which is x=43 TL. We insert this value in the price so that p(43)=150-2·43=64 TL.

The equilibrium point of given demand and supply functions is the point (4,5). Therefore, we must look for this point for graphs. Correct answer is B.  

The price of a product at the intersection point, provided that it is in the first quadrant, is called market price. Equate the supply and demand function giving;

x is found as 30. When we put it on the demand or supply equation, we find market price as 135. 

f(x)=x²+lnx-1

f'(x)=2x+(1/x)

f''(x)=2-(1/x²)

If we substitute x=0 and y=0 the result is 0/0 which is undefined. So we will arbitrarirly take y=kx.

Then:

(x²+3y²)/(x²-3y²)=(x²+3k²x²)/(x²-3k²x²)=x²(1+3k²)/x²(1-3k²)=(1+3k²)/(1-3k²)

The result depends on the value of k. For instance if k=2 result is -13/11, but if k=4 result is -49/47.

Therefore the limit does not exit.