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

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

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

$z_x=\dfrac{2xy^3}{\cos z-z\sin z-1}$

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

$\nabla f(x,y,z)=(2xz+y^2,2xy+z^2,2yz+x^2)$

and the gradient at the desired point is

$\nabla f(-1,1,2)=(-3,2,5)$

The tangent plane then has equation

$\nabla f(-1,1,2)\cdot(x+1,y-1,z-2)=0$

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

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

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

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).