A Scalene triangle has all of its three lengths to be different. The traingle above can not be an equilateral triangle too because all the sides aren't equal.

Also, it's not an isosceles triangle since two of its sides are not thesame. It's not a right angle triangle as the square of the hypothenuse isn't equal to the addition of the square of the other two sides.

6 0

3 years 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.

4 0

4 years ago

mustafa is putting a sidewalk using two diffrent style bricks, one style brick is 8 inches long, and the other style brick is 6

Svetlanka [38]

Answer:

8y+6y=288

Step-by-step explanation:

8y+6y= 12y

12y=288

12y divide by 12 = 288 divided by 12 =

y=24

3 0

3 years ago

A sample contains 1665 bacteria. The bacteria decreases at a rate of 13% per day.​

A sample contains 1665 bacteria. The bacteria decreases at a rate of 13% per day.