Find two numbers x and y such that a) 2x+y=100 and A=2x+2xy+y is maximized b) 2x+4y-15=0 and B= √x2+y2is minimized. Note that in this case, minimizing B is equivalent to minimizing B2.

a) x = 25, y = 50

b) x = 1.5, y = 3

Step-by-step explanation:

We have to use Lagrange Multipliers to solve this problem. The maximum of a differentiable function f with the constraint g(x,y) = b, then we have that there exists a constant such that

Or, in other words,

a) Lets compute the partial derivates of f(x,y) = 2x+2xy+y. Recall that, for example, the partial derivate of f respect to the variable x is obtained from derivating f thinking the variable y as a constant.

On the other hand,

The restriction is g(x,y) = 100, with g(x,y) = 2x+y. The partial derivates of g are

This means that the Lagrange equations are

Note that 2y + 2, which is is the double of 2x+1, which is . Therefore, we can forget for now and focus on x and y with this relation:

2y+2 = 2 (2x+1) = 4x+2

2y = 4x

y = 2x

If y is equal to 2x, then

g(x,y) = 2x+y = 2x+2x = 4x

Since g(x,y) = 100, we have that

4x = 100

x = 100/4 = 25

And, therefore y = 25*2 = 50

Therefore, x = 25, Y = 50.

b) We will use the suggestion and find the minumum of f(x,y) = B² = x²+y², under the constraing g(x,y) = 0, with g(x,y) = 2x+4y-15. The suggestion is based on the fact that B is positive fon any x and y; and if 2 numbers a, b are positive, and a < b, then a² < b². In other words, if (x,y) is the minimum of B, then (x,y) is also the minimum of B² = f.

Lets apply Lagrange multipliers again. First, we need to compute the partial derivates of f:

And now, the partial derivates of g:

This gives us the following equations:

If we compare 2x with 2y, we will find that 2y is the double of 2x, because 2y is equal to , while on the other hand, . As a consequence, we have

2y = 2*2x

y = 2x

Now, we replace y with 2x in the equation of g:

0 = g(x,y) = 2x+4y-15 = 2x+4*2x -1x = 10x-15

10 x = 15

x = 15/10 = 1.5

y = 1x5*2 = 3

Then, B is minimized for x 0 1.5, y = 3.

12/11c

or

12

c

11

the results from a pre-test for students for the year 2000 and the year 2010 are illustrated in the box plot. what do these results tell us about how students performed on the 29 question pre-test for the two years?

if we compare only the lowest and highest scores between the two years, we might conclude that the students in 2010 did better than the students in 2010. this conclusion seems to follow since the lowest score of 8 in 2010 is greater in value than the lowest score of 6 in 2000. also, the highest score of 28 in 2010 is greater in value than the highest score of 27 in 2000.

but the box portion of the illustration gives us more detailed information. the middle bar in each box shows us that the median score of 20 in 2000 is greater in value than the median score of 17 in 2010. further, we note that the box and whiskers divide the illustration into four pieces. each of these four pieces represents the same portion of students. so, the upper half of the students in 2000 scored in the same score range as the upper one-fourth of the students in 2010, see the illustration at a score of 20.

by considering the upper one-fourth, upper half, and upper three-fourths instead of just the lowest and highest scores, we would conclude that the students as a whole did much better in 2000 than in 2010. we would conclude that as a whole the students in 2010 are less prepared than the students in 2000.

in this section, we discuss box-and-whisker plots and the five key values used in constructing a box-and-whisker plot. the key values are called a five-number summary, which consists of the minimum, first quartile, median, third quartile, and maximum.

step-by-step explanation: