Identify the graphed linear equation.
2 answers:
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Answer:
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
(this is the restriction, in other words, g(x,y) = 100)
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.