Y is inversely proportional to x . When y = 3 , x = 20 Work out x when y = 12

Mathematics

1 answer:

Talja [164]2 years ago

5 0

Answer:

x = 5 when y = 12

Step-by-step explanation:

If y is inversely proportional to x, we can say that x1 : x2 = y2 : y1

The order of y2 and y1 respectively are interchanged as y is inversely proportional to x

Now let us simply substitute our known values, and solve for x2;

x1 : x2 = y2 : y1

20 : x2 = 12 : 3

20/x2 = 12/3

12(x2) = 3(20)

12(x2) = 60

x2 = 60/12 = 5

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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.

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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.