Please fast, please fast

Find the absolute maximum and minimum values of f(x,y)=xy−4x in the region bounded by the x-axis and the parabola y=16−x2.

skad [1K]

Answer:

The absolute maximum and minimum is $20\; \text{and} -20.$

Step-by-step explanation:

We first check the critical points on the interior of the domain using the

first derivative test.

$f_x=y-4=0$

$f_y=x=0$

The only solution to this system of equations is the point (0, 4), which lies in the domain.

$f_{xx}=0, \;f_{yy}=0\; \text{and}\; f_{xy}=-1$

$\Rightarrow f_{xx}f_{yy}-f_{xy}=o-1=-1$

$\therefore (0,4)$ is a saddle point.

Boundary points - $(4,0), (-4,0), (0,16)$

Along boundary $y=16-x^2$

$f=x(16-x^2)-4x$

$=16x-x^3-4x$

$\Rightarrow f^'=16-3x^2-4=0$

$\Rightarrow 3x^2=12$

$\Rightarrow x=\pm2,\;\;y=14$

Values of f(x) at these points.

$\begin{array}{}(4,0)=-16\\(-4,0)=16\\(0,16)=0\\(2,14)=20\\(-2,14)=-20\end{array}$

Therefore, the absolute maximum and minimum is $20\; \text{and} -20.$

6 0

3 years ago

Simplify: 9i(2+3i) enter your answer, in standard form, in the box ___

harina [27]

-54......................

8 0

3 years ago

Read 2 more answers

Please help me with this question please

zheka24 [161]

*im sorry but I cant see the bottom number but I'll consider it as x

tan 45° = 3/x

x = 3/tan45

x is 3

hope this helps :)

5 0

3 years ago

PLease help me LOTS of points

notsponge [240]

Answer:

15) a) 8

b) y + 2

c) y - 1

16) True

17) 2n - 1, where n is an integer

18) a) 3p + 4q

b) 500 - 3p - 4q

4 0

3 years ago

The figure shows a construction completed by hand.

Tresset [83]

Based on the construction, we can logically deduce that: A. yes; the compass was kept at the same width to create the arcs for points C and D.

<h3>What is a line segment?</h3>

A line segment can be defined as the part of a line in a geometric figure such as a triangle, circle, quadrilateral, etc., that is bounded by two (2) distinct points and it typically has a fixed length.

In Geometry, a line segment can be measured by using the following measuring instruments:

A scale (ruler)

A divider

A compass

In Geometry, an arc can be defined as a trajectory that is generally formed when the distance from a given point has a fixed numerical value.

Based on the construction with arcs created above and below the line segment from points A, we can infer and logically deduce that it is true that the compass was kept at the same width to create the arcs for points C and D.

In conclusion, yes, the construction demonstrated how to bisect a line segment correctly by hand.

Read more on arcs here: brainly.com/question/11126174

#SPJ1

Complete Question:

The construction has a given segment AB. Arcs have been created above and below the segment from points A that are equidistant from point A. The compass was kept at the same distance, placed on point B, and two additional arcs were created above and below the segment that intersect with the first arcs created. The intersection of the arcs above the segment created point C. The intersection of the arcs below the segment created point D. A line was drawn from point C to D through the segment.

Does the construction demonstrate how to bisect a segment correctly by hand?

A. Yes; the compass was kept at the same width to create the arcs for points C and D.

B. Yes; a straightedge was used to create segment CD.

C. No; the compass was not kept at the same width to create the arcs for points C and D.

D. No; a straightedge was used to create segment CD.

7 0

2 years ago