How to find abs max and min of f(x,y)

Bounded region D={(x,y): 0<=x<=pi, 0<=y<=pi }
f(x,y)= 3xy + 2cos(x)

How do you find the abs. max and min values for the above function?

First, we find the critical point inside D.

f_x = 3y - 2 sin x, f_y = 3x.

Setting these equal to 0:
3y - 2 sin x = 0
3x = 0 ==> x = 0.

So, 3y - 0 = 0 ==> y = 0.

So, we have the point (0, 0) [which is on the boundary of D].
Note that f(0, 0) = 2.
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Next, we check the boundary, one edge at a time.

(i) x = 0 with y in [0, π]:
==> f(0, y) = 0 + 2 = 2, which is constant.

(ii) x = 0 with y in [0, π]:
==> f(π, y) = 3πy - 2.
This takes its extrema at the endpoints:
f(π, 0) = -2 and f(π, π) = 3π^2 - 2.

(iii) y = 0 with x in [0, π]:
==> f(x, 0) = 2 cos x
This takes its extrema at the endpoints:
f(0, 0) = 2 and f(π, 0) = -2.

(iv) y = π with x in [0, π]:
==> g(x) := f(x, π) = 3πx + 2 cos x.

g'(x) = 3π - 2 cos x > 0 for all x (since cosine is bounded between -1 and 1 inclusive).
So, g is increasing for all x.

This again takes its extrema at the endpoints:
f(0, π) = 2 and f(π, π) = 3π^2 - 2.
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So, the absolute extrema are -2 and 3π^2 - 2.

I hope this helps!

Aye!!

g'(x) = 3π - 2 sin x > 0 since sine is between -1 and 1 inclusive; so all remarks above about g and g still apply

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TY!

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