Question

Find the absolute minimum value of the function f(x,y) = 6 + 3xy -2x- 4y on the set D. D is bounded by the parabola y=x^2 and y=4

Answer 1

If you compute the gradient of f, you find it
gradf={-2 + 3 y, -4 + 3 x}
You find that this gradient is zero outside D. Hence any minimum and maximum must occur at the boundary of D. The problem is then reduced to finding the extrema of
f(x,x^2) and f(4,x) for x between 2 and -2
Can you finish it now?

Answer 2

I've never being taught to find the abs. value with the gradients, only directional derivative...

Answer 3

I've being taught using my partial derivative, ( which I understand is like a gradient). But to me a griadent is a vector.

Answer 4

It is a vector to me too. I wrote its two components.

Answer 5

I understand what you did to find the gradient, but how you used it to establish that the boundaries must lie at the limit of the set... I've never encountered that, I was always given sets that were numerical ( so 0</= x </= 3 for instance)

Answer 6

hence why I'm thrown off, when I sketch my bounded region ( usually a rectangle or square) I don't know how to use the method I was instructed which was to divided it into 4 different segment and solve for each one.

Answer 7

Here your boundary has two parts, the parabola y=x^2 and the line y=4 that are part of D

Answer 8

Any point on the parabola can be written ans (x,x^2) and any point on the line can be written (x,4)

Answer 9

After you analyse the situation, you find that the absolute minimum is
f(-2,4)=-30
and the absolute maximum is
f(2,4)=10

Answer 10

but, how do you go about finding the absolute minimum, this is where I don't know what to do. I found my critical point, but then how do you establish the "boundaries" of x?

Answer 11

first you need to find the critical point, by setting fx = 0, and fy = 0
Then make sure that the critical point is inside the region. Then you check for boundary

Answer 12

why does it matter if the critical point is inside the region? ( just by curiosity) and I did find it .. 2/3 and 4/3

Answer 13

because if the critical point is outside of the region, then you can only obtain the absolute max/min along the boundary

Answer 14

the critical point is inside the region, or so it looks like? if i draw it out its bounded from (-2,4) to (2,4) and (2/3, 4/3) is in the region. Or so I think

Answer 15

yes, it's inside the region. Now you need to check the boundary

Answer 16

and how do I do so, I was taught checking the boundaries by doing the following:
f(x,x^2) ---> for y=x^2
so f(x,x^2)= 6+3x^2-2x+4x^2

Answer 17

but i don't know what to do with that.

Answer 18

find the derivative and set it equal to 0, also make sure the critical point is along the region. Do the same when y = 4

Answer 19

here are the points you need to evaluate;
f(4/3, 2/3)
f(0.623, 4.652)
f(-0.178, 6.195)
f(-2,4)
f(2,4)

whichever is the largest is the maximum, whichever is the smallest is the minimum

Answer 20

how did you solve for those points, that is where I am at lost.

Answer 21

well how did you know they represented a critical point.

Answer 22

(4/3, 2/3) is outside the region.

Answer 23

sorry the actual critical points were (2/3,4/3)

Answer 24

f(x,x^2) is a function of one variable, so you find its absolute min and max in the usual way. Here also it happened that they occur at the end points x=-2 and x=2

Answer 25

No the critical points are (4/3, 2/3). What make you think otherwise?

Answer 26

What makes the gradient vector gradf={-2 + 3 y, -4 + 3 x} ={0,0}
-2 + 3 y =0
-4 + 3 x=
Solve these two equations and you find y=2/3 and x=4/3

Answer 27

-4 + 3 x=0
This was a misprint in my last post