MATEMATİK I - Deneme Sınavı - 3

(5x²+k²x²) / (x²+k²x²)=x²(5+k²) / x²(1+k²)=(5+k²) / (1+k²).

By the way we have an answer depending on k, the limit does not exist.

∂f/∂x=(x.lnx)'=x'.lnx+x.lnx'

=lnx+1

∂f /∂x=6xy+ye^xy

^{∂²f /∂x²=6y+y²}e^xy

∂f / ∂y = [(2x²y).(x+y)-(x²y²).1] / (x+y)²

By the way x=1,y=2

∂f / ∂y = [(4).(3)-(4).1] / (3)²=8/9

$\partial f/\partial x\; =2/z=ln(2x+y)">(2x+y)\; =2$

$z=ln(2x+y)">\partial f/\partial y=1/z=ln(2x+y)">(2x+y)=1$

$z=ln(2x+y)">then,\; z-0=2(x-(-1))+1(y-3)$

$z=ln(2x+y)">z=2(x+1)+y-3$

$z=ln(2x+y)">z=2x+y-1$

∂z / ∂v = (2x+y) . (-1) + (x) . (-u)

Hence, x=u²-v and y=-uv

∂z / ∂v = (2(u²-v)+(-uv)) . (-1) + (u²-v) . (-u)

By the way u=2, v=-1

∂z / ∂v = (2(5)+(2)) . (-1) + (5) . (-2)= -22

∂z / ∂v = (∂z / ∂x) . (∂x / ∂v)+ (∂z / ∂y) . (∂y / ∂v)

∂z / ∂v = (2x+(y/x)) . (2v) + (lnx) . (1)

Hence, x=u³+v² and y=v

∂z / ∂v = (2(u³+v²)+(v/(u³+v²))) . (2v) + (ln(u³+v²)) . (1)

On substitution u=1, v=0

∂z / ∂v = (2(1)+(0)) . (0) + (0) . (1)= 0

for the critical points;

4x+ 2y-6=0 and  2x+4y=0, so x=2, y=-1.

∂²f / ∂x² =4         ∂²f / ∂y² =4       ∂f / ∂xy =2

D=4.4-2²=12.

Since D>0, the function has a local minimum at (2,-1).

L(x, y, λ) = x+2y+ λ(xy-6050).

∂L/∂x =0, ∂L/∂y =0, ∂L/∂λ =0

1+λy=0, 2+λx=0, xy-6050=0

Hence, λ=-1/y=-2/x

x=2y

2y²=6050

y=55, x=110

x+2y=220 m.