MATEMATİK I

x²-4y²≥0

(x-2y)(x+2y)≥0

This can be true only if both (x-2y) and (x+2y) are non-negative or both are non-positive.

I. First we take the first case (both are non-negative).

Thus:

x-2y≥0 which means x≥2y

x+2y≥0 which means x≥-2y

Thus the solution set is x≥2y ∩ x≥-2y

II. Now we take the second case (both are non-positive)

Thus:

0≥x-2y which means 2y≥x

0≥x+2y which means -2y≥x

This can only be true when x=y=0, so the answer is , the answer is A

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.

f_x=10x+y (since the derivative of y² is zero)

Since z=(x+y)², dz/dx=2x+2y and dz/dy=2x+2y 

Since y=u³, dy/du=3u²

x=u²-u, dx/du=2u-1

By substitution

dz/du=(2x+2y)(2u-1)+(2x+2y)3u²=(2x+2y)(3u²+2u-1)=(2(u²-u+3u³)(3u²+2u-1))=(2u²-2u+3u³)(3u²+2u-1)=6u⁵+10u⁴-4u³-6u²+2u

2xydx+x²dy-6ydy+2xdz+2zdx=0

Since we are interested in dz/dx we can take dy=0 (y doesnt is constant).

So 2xydx+2xdz+2zdx=0 then 2xdz=-2xydx-2zdx and divind both sides to dx we obtain 2xdz/dx=-2xy-2z

so dz/dx=(-2xy-2z)/2x=-y-z/x

Since 120=2x+4y, y=(120-2x)/4=30-0.5x

Then we can rewrite his utility function as U(x,y)=x^0.5y^0.5=U(x)=x^0.5(30-0.5x)^0.5=(x(30-0.5x))^0.5=(30x-0.5x²)^0.5

For the maximum utility the first derivative of utility with respect to x must be equal to zero. Thus:

Ux=0.5(30-x)(30x-0.5x²)^-0.5=0

30-x=0, which means x=30. Substituting this in y=30-0.5x=15. Thus the utility maximizing combination of (x,y)=(30,15)

We have to find the points simultaneously satisfy fx=0 and fy=0

fx=4x³+2xy²-4x=0, so x(4x²+2y²-4)=0

fy=2x²y+4y=0, so y(2x²+4)=0

Points given in III and V doesnt satisfy these conditions simultaneously. So the remaining 3 points are stationary points.

2y+2x=40, y=20-x

f(x,y)=g(x)=x(20-x)=20x-x² will be the function of one variable to be maximized.

g_x=20-2x=0 then x=10, since y=20-x=10. So the area will be 10*10=100