If (-1, -1) is an extremum of , then both partial derivatives vanish at this point.
Compute the gradients and evaluate them at the given point.
The first and third functions drop out.
The second function depends only on . Compute the second derivative and evaluate it at the critical point .
This indicates a minimum when . In fact, since this function is independent of , every point with this coordinate is a minimum. However,
for all , so (-1, 1) and all the other points are actually <em>global</em> minima.
For the fourth function, check the sign of the Hessian determinant at (-1, 1).
The second derivative with respect to is -2/(-1) = 2 > 0, so (-1, -1) is indeed a local minimum.
The correct choice is the fourth function.
Answer:
Step-by-step explanation:
Let
x ----> the width of the rectangular corral
y ----> the length of the rectangular corral
we know that
The perimeter of the rectangular corral is equal to
so
simplify
----> equation A
Remember that
The area of a rectangle is equal to
----> equation B
substitute equation A in equation B
Convert to function notation
We use the distance formula for this problem.
d = √[(x₂ - x₁)² + (y₂ - y₁)²]
The distance between point (-2,-2) and point (-2,4).
d = √[(⁻2 - ⁻2)² + (4 - ⁻2)²] = 6 units
Then, compute for 20% of 6 units:
Distance traveled = 6(0.2) = 1.2 units
Use 1.2 units as distance and the starting point (-2,-2). The x-coordinate should still be at -2 because the distance is a straight line as shown in the picture.
1.2 = √[(-2 - ⁻2)² + (y - ⁻2)²]
Solving for y,
y = -0.8
The point is found at (-2,-0.8). This is located at quadrant 3. As to the distance traveled, that would be: 1.2*6 = 6 miles. Thus, the answer is C.
And the reason is that for most purposes we can obtain suitable accuracy quickly and inexpensively on information gained from a sample. The bottom line is it would be wasteful and foolish to use the entire population when a sample, drawn scientifically, provides accuracy in representing your population of interest.