Calculus 3 question: Find the absolute maximum and absolute minimum of f(x,y)?

Question

f(x,y)= 2+2x+2y-x^2-y^2 over the triangular region with vertices(0,0), (9,0), (0,9).


How do I attempt to solve this problem?
thanks in advance

Answer 1

When you maximize or minimize over an area, the critical points (maxima and minima) will arise either where df/dx and df/dy are both zero, or on vertices, or on the boundaries.



We first set df/dx = 0 and dg/dx = 0.

df/dx = 2 - 2x = 0 => x = 1

dg/dx = 2 - 2y = 0 => y = 1

Thus, one critical point is (1,1). (We do not yet know if this is a maxima or a minima, but lies in the region)



Next, we consider the three boundary lines. These are x = 0, y = 0, and x + y =9.



For x = 0, f(x,y) = 2 + 2y - y^2. We can directly find the critical points by taking

df/dy = 2 - 2y = 0 => y = 1. This yields another critical point (0,1)



For y = 0, f(x,y) = 2 + 2x - x^2. We can directly find the critical points by taking

df/dx = 2 - 2x = 0 => x = 1. This yields another critical point (1,0)



For x + y = 9, we can use Langrange multipliers.

Define h(x,y,L) = f(x,y) + L (x + y -9)

The critical points will be obtained by setting dh/dx = dh/dy = dh/dL = 0. We have:

dh/dx = 2 - 2x + L = 0

dh/dy = 2 - 2y + L = 0

dh/dL = x + y - 9 = 0



The first two equations yield x = y. The last equation gives x = y = 9/2.

So the next critical point is (9/2, 9/2).





Finally, all three vertices (0,0), (0,9), (9,0) are also critical points.



Thus, the set of all critical points is {(1,1), (1,0), (0,1), (9/2,9/2), (0,0), (9,0), (0,9)}.

Check the value of f(x,y) at all these points. It is maximum at (1,1) and equals 4.

It is minimum at (9,0) and (0,9) and equals -61.

Answer 2

-(x^2 + y^2 -2x-2y+2-2-2)=f(x,y)

(x-1)^2 + (y-1)^2 + 4 = f(x,y)

put y-1=p and x-1 =q

this becomes : p^2 + q^2 - 4 = f(p,q)

now find df/dp and df/dq and check on vertices first ull get ur maxima at (9/2,9/2) and minima at (0,0)

just plot5 and region and the curve ull c for itself

hope it helps

Answer 3

Why would not zero be an absolute minimal importance? Look on the graph for 4x^three - it has no highest or minimal importance The restrict operator tells us that as x-> -? then 4x^three is going to -? x-> +? then 4x^three is going to +? This position does now not have a min or max, however it does have an inflection factor at (zero,zero) Look on the graph for 3x^four : it is a parabola with a minimal importance. Whoops, it is fairly { - (3x^four) } so it is a parabola with a highest importance. This parabola opens downward and its highest might be at (zero,zero) The restrict operator tells us that as x-> -? then 3x^four is going to -? x-> +? then 3x^four is going to -? ===== Adding any position bounded on a truly quantity and further to one more position that has infinities will nonetheless influence in a position that has infinities on the identical limits. x^four is regularly going to be a confident quantity for all values of x < zero x^three is regularly going to be a poor quantity for all values of x < zero So four*x^three will regularly be poor As three*x^four will regularly be confident Then 4x^three-3x^four will regularly be a poor quantity for x < zero and it'll lessen as x decreases, use the restrict operator to look this. =============

Answer 4

derivate it ie find dy/dx and name it suppose f'(x),,put this as 0 and find the value of x(s) ie if there r two values for dat eqn..then find the derivate of f'(x) ied(f'(x)/dx name it f''(x).now put the previously found x's value in dis..if it comes <0,the function is max.for dat value of x and min.for a value of f''(x)>0..hope ive helped..:)