Find the absolute maximum and minimum values of f on the set D.
28. f(x, y) = 3 + xy ? x ? 2y, D is the closed triangular region with vertices (1, 0), (5, 0), and (1, 4)
24. Use Lagrange multipliers to prove that the triangle with maximum area that has a given perimeter p is equilateral.
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From the matrix of 2nd derivatives of f(x,y) we see that is does not produce a sign-defined quadratic form, therefore f(x,y) has no local extrema anywhere, and so its absolute extrema in a bounded region ar on the boundaries of the region.
In this case these are the sides of the triangle.
The side (1,0)-(5,0) has y = 0 and so f = 3 - x on it, with extrema at the vertices (1,0) and (5,0), equal to 2 and -2, respectively.
The side (1,0)-(1,4) has x = 1 and so f = 2 - y on it, with extrema at the vertices (1,0) and (1,4), equal to 2, and -2, respectively.
The side (5,0)-(1,4) has y = 5 - x and so f = -x^2+6x-7 = -(x-3)^2 + 2, with maximum at x=3 (y=2), equal to 2.
Therefore the absolute minimum and maximum of f(x,y) on this triangle are -2 and 2, respectively.
There is a Heron-Archimedes formula for the area of a triangle in terms of its sides (if you do not have it in your books, you can use for reference web page http://www.efunda.com/math/areas/triangleGen.cfm).
It will be convenient for us to re-write it in terms of 16 times the square of its area:
where A is the area, a,b, and c are the sides, and p = a+b+c is the perimeter.
Now we want to maximize the area for a constant perimeter, and so introduce a Lagrangian multiplier t, and maximize the
G(a,b,c,t) = p(p-2a)(p-2b)(p-2c) - t(p - a - b - c) (2)
Equating to zero the partial derivatives of G with respect to a,b, and c, we get equations
t = 2p(p-2b)(p-2c) (3a)
t = 2p(p-2a)(p-2c) (3b)
t = 2p(p-2a)(p-2b) (3c)
Now, if we divide the left and right sides of (3a) by respective sides of (3b), we obtain
p-2b = p-2a, so that a = b,
and doing the same for the pair (3a)/(3c), we obtain
p-2c = p-2a, so that a = c.
Therefore, a=b=c for the area to be maximal for a given perimeter.