Question

https://hizliresim.com/ID0Ydx

can you explain these questions in detail? By writing all the details on the paper for each question. Thanks ..:)

Answer 1

1. Assuming
`z=z(x,y)`, we have

`D_x(z\cos z)=z_x\cos z-z\sin z\,z_x=z_x(\cos z-z\sin z)`

`D_x(x^2y^3+z)=2xy^3+z_x`

so that

`z_x(\cos z-z\sin z)=2xy^3+z_x`

`z_x(\cos z-z\sin z-1)=2xy^3`

`z_x=(2xy^3)/(\cos z-z\sin z-1)`

2. Let
`f(x,y,z)=zx^2+xy^2+yz^2`. The gradient of
`f` is

`\\abla f(x,y,z)=(2xz+y^2,2xy+z^2,2yz+x^2)`

and the gradient at the desired point is

`\\abla f(-1,1,2)=(-3,2,5)`

The tangent plane then has equation

`\\abla f(-1,1,2)\cdot(x+1,y-1,z-2)=0`

`(-3,2,5)\cdot(x+1,y-1,z-2)=0`

`-3(x+1)+2(y-1)+5(z-2)=0`

`-3x+2y+5z=15`

3.
`f` has critical points where the derivatives vanish or do not exist. The latter is irrelevant here, since
`f` is a polynomial and is thus continuous everywhere. So just differentiate and set the derivative equal to 0:

`f_x=y^2-2xy^2+4x^3=0`

`f_y=2xy-2x^2y=0\implies 2xy(1-x)=0`

The last equation tells us
`x=0`,
`y=0`, or
`x=1`.

If
`x=0`, then in the first equation we get
`y^2=0\implies y=0`.

If
`y=0`, then
`4x^3=0\implies x=0`.

If
`x=1`, then
`y^2-2y^2+4=0\implies y^2=4\implies y=\pm2`.

So there are 3 critical points at (0, 0), (1, -2), and (1, 2).

4. Stationary points are critical points where the derivative vanishes. So same process as in (3):

`f(x,y)=xye^(x+y)`

`f_x=ye^(x+y)+xye^(x+y)=0\implies ye^(x+y)(1+x)=0`

`f_y=xe^(x+y)+xye^(x+y)=0\implies xe^(x+y)(1+y)=0`

The first equation tells us
`y=0` or
`x=-1`.
`e^(x+y)>0` for all
`x,y`, so we can ignore this term.

If
`y=0`, then
`xe^x=0\implies x=0`.

If
`x=-1`, then
`-e^(y-1)(1+y)=0\implies y=-1`.

The second equation tells us
`x=0` or
`y=-1`. But this gives the same information as before.

So there are only two stationary points, (0, 0) and (-1, -1).

5. First find the stationary points:

`f_x=2x+y=0`

`f_y=x+2y=0`

Solve this to find the one point at (0, 0). At this point, we have
`f(0,0)=0`.

Now check for extrema along the boundary. The rectangle is bounded by the lines
`x=-2`,
`x=2`,
`y=-1`, and
`y=1`.

On
`x=-2`, we have

`f(-2,y)=y^2-2y+4=(y-1)^2+3`

which attains a minimum of 3 when
`y=1` and a maximum of 7 when
`y=-1`.

On
`x=2`, we have

`f(2,y)=y^2+2y+4=(y+1)^2+3`

which has a minimum of 3 at
`y=-1` and a maximum of 7 at
`y=1`

We can check on the other 2 lines, but we'd get the same information.

So we've found that
`f` has an absolute minimum of 0 at (0, 0), and an absolute maximum of 7 at both (-2, -1) and (2, 1).