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const2013 [10]
3 years ago
11

Which equation matches this table?

Mathematics
1 answer:
alekssr [168]3 years ago
8 0

9514 1404 393

Answer:

  y = 6x

Step-by-step explanation:

When x increases by 1 (from 1 to 2), y increases by 6 (from 6 to 12). So, none of the addition equations will work to express this relationship. The appropriate choice is ...

  y = 6x

A test for other x-values reveals this explains all of the y-values shown.

<em>Example</em>:

  x = 7, y = 6·7 = 42

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To prove that jill is wrong we just need an example of this;


2*3*5*7*11*13 = 30030 (this is the smallest number with 6 different prime numbers)

5953*5981*5987 = 2.13x10^11 (which is obviously a much bigger number)

this is enough to prove that jill is wrong
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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.
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.
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