Find the slope and the y-intercept of each line from it's equation. 3x-y=28

Mathematics

1 answer:

Otrada [13]2 years ago

7 0

<span>
Slope = 3

x-intercept = 28/3 = 9.33333<span>

y-intercept = 28/-1 = -28.00000

Hope this helped</span></span>

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bija089 [108]

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1 year ago

Find all points on the portion of the plane x+y+z=5 in the first octant at which f(x, y, z) = xy2z2 has a maximum value.

irina1246 [14]

Lagrange multipliers:

$L(x,y,z,\lambda)=xy^2z^2+\lambda(x+y+z-5)$

$L_x=y^2z^2+\lambda=0$
$L_y=2xyz^2+\lambda=0$
$L_z=2xy^2z+\lambda=0$
$L_\lambda=x+y+z-5=0$

$\lambda=-y^2z^2=-2xyz^2=-2xy^2z$

$-y^2z^2=-2xyz^2\implies y=2x$ (if $y,z\neq0$ )

$-y^2z^2=-2xy^2z\implies z=2x$ (if $y,z\neq0$ )

$-2xyz^2=-2xy^2z\implies z=y$ (if $x,y,z\neq0$ )

In the first octant, we assume $x,y,z>0$ , so we can ignore the caveats above. Now,

$x+y+z=5\iff x+2x+2x=5x=5\implies x=1\implies y=z=2$

so that the only critical point in the region of interest is (1, 2, 2), for which we get a maximum value of $f(1,2,2)=16$ .

We also need to check the boundary of the region, i.e. the intersection of $x+y+z=5$ with the three coordinate axes. But in each case, we would end up setting at least one of the variables to 0, which would force $f(x,y,z)=0$ , so the point we found is the only extremum.