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harina [27]
3 years ago
8

When 'x' and 'y' are in direct proportion , then the relation between 'x' and 'y' is __________

Mathematics
1 answer:
Bess [88]3 years ago
5 0

Step-by-step explanation:

If two quantities x and y vary (change) together in such a manner that the ratio of their corresponding values remains constant, then x and y are said to be in direct proportion.

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3 years ago
The plane x + y + 2z = 12 intersects the paraboloid z = x2 + y2 in an ellipse. Find the points on the ellipse that are nearest t
Soloha48 [4]

The distance between a point (x,y,z) and the origin is \sqrt{x^2+y^2+z^2}. But since (\sqrt{f(x)})'=\frac{f'(x)}{2\sqrt{f(x)}}, both f(x) and \sqrt{f(x)} have the same critical points, so we can consider instead the squared distance, x^2+y^2+z^2.

We're looking for the extrema of x^2+y^2+z^2 subject to x+y+2z=12 and z=x^2+y^2. The Lagrangian is

L(x,y,z,\lambda,\mu)=x^2+y^2+z^2+\lambda(x+y+2z-12)+\mu(z-x^2-y^2)

with critical points where the partial derivatives vanish:

L_x=2x+\lambda-2\mu x=0\implies\lambda=2x(\mu-1)

L_y=2y+\lambda-2\mu y=0\implies\lambda=2y(\mu-1)

L_z=2z+2\lambda+\mu=0

L_\lambda=x+y+2z-12=0

L_\mu=z-x^2-y^2=0

From the first two equations, it follows that x=y.

Then in the last two equations,

x+y+2z-12=0\implies x+z=6

z-x^2-y^2=0\implies z=2x^2

\implies x+2x^2=6\implies2x^2+x-6=(2x-3)(x+2)=0\implies x=\dfrac32\text{ or }x=-2

If x=\frac32, then z=6-\frac32=\frac92; if x=-2, then z=8.

So there are two critical points, \left(\frac32,\frac32,\frac92\right) and (-2,-2,8).

Let f(x,y,z)=\sqrt{x^2+y^2+z^2}. We have a minimum distance of f\left(\frac32,\frac32,\frac92\right)=\boxed{\frac{3\sqrt{11}}2} and maximum distance of f(-2,-2,8)=\boxed{6\sqrt2}.

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4 years ago
Quiz 3
andrezito [222]
It is 2x+8!!!!!!! I’m pretty sure
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