Victoria had $200 in her account at the end of one year. At the first of each subsequent year she deposits $15 into
2 answers:
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Answer:
Step-by-step explanation:
Let:
This is an exact differential equation because:
With this in mind let's define f(x,y) such that:
and
So, the solution will be given by f(x,y)=C1, C1=arbitrary constant
Now, integrate with respect to x in order to find f(x,y)
where g(y) is an arbitrary function of y
Let's differentiate f(x,y) with respect to y in order to find g(y):
Now, let's replace the previous result into :
Solving for
Integrating both sides with respect to y:
Replacing this result into f(x,y)
Finally the solution is f(x,y)=C1 :
Answer:
a)
Highest (-3,-3)
Lowest (2,2)
b)
Farthest (-3,-3)
Closest (2,2)
Step-by-step explanation:
To solve this problem we will be using Lagrange multipliers.
a)
Let us find out first the restriction, which is the projection of the intersection on the XY-plane.
From x+y+z=12 we get z=12-x-y and replace this in the equation of the paraboloid:
completing the squares:
and we want the maximum and minimum of the paraboloid when (x,y) varies on the circumference we just found. That is, we want the maximum and minimum of
subject to the constraint
Now we have
Let be the Lagrange multiplier.
The maximum and minimum must occur at points where
that is,
we can assume (x,y)≠ (-1/2, -1/2) since that point is not in the restriction, so
Replacing in the constraint
from this we get
<em>x=-1/2 + 5/2 = 2 or x = -1/2 - 5/2 = -3 </em>
<em> </em>
and the candidates for maximum and minimum are (2,2) and (-3,-3).
Replacing these values in f, we see that
f(-3,-3) = 9+9 = 18 is the maximum and
f(2,2) = 4+4 = 8 is the minimum
b)
Since the square of the distance from any given point (x,y) on the paraboloid to (0,0) is f(x,y) itself, the maximum and minimum of the distance are reached at the points we just found.
We have then,
(-3,-3) is the farthest from the origin
(2,2) is the closest to the origin.