Find the maximum value M of the function f(x,y) = x^7y^5(3−x−y)^4 on the region x ≥ 0, y ≥ 0, x + y ≤ 3. M=

Find the maximum value M of the function f(x,y) = x^7y^5(3−x−y)^4
on the region x ≥ 0, y ≥ 0, x + y ≤ 3.

M=...........

First, find the critical points inside the region.

f_x = 7x^6 y^5 * (3 - x - y)^4 + x^7 y^5 * -4(3 - x - y)^3
.....= x^6 y^5 (3 - x - y)^3 [7(3 - x - y) - 4x]
.....= x^6 y^5 (3 - x - y)^3 (21 - 11x - 7y)

f_y = 5x^7 y^4 * (3 - x - y)^4 + x^7 y^5 * -4(3 - x - y)^3
.....= x^7 y^4 (3 - x - y)^3 [5(3 - x - y) - 4y]
.....= x^7 y^4 (3 - x - y)^3 (15 - 5x - 9y).

Set these equal to 0:
x^6 y^5 (3 - x - y)^3 (21 - 11x - 7y) = 0
x^7 y^4 (3 - x - y)^3 (15 - 5x - 9y) = 0.

So, x = 0, y = 0, or 3 - x - y = 0 (these are the boundaries of the region; so ignore them).
==> 21 - 11x - 7y = 0, and 15 - 5x - 9y = 0.
==> (x, y) = (21/16, 15/16), which is inside the region.

Note that f(21/16, 15/16) = 1.54 (approx.)
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Now, we check the boundary lines of the region, one at a time.
(i) x = 0 or y = 0 ==> f(x,y) = 0.

(ii) x + y = 3 <==> y = 3 - x with x in [0, 3].
==> f(x, y) = 0.

Hence the maximum value is 1.54.

I hope this helps!