# Absolute Minimum and Maximum Value of a Function on a Set

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.

## Solution This solution is **FREE** courtesy of BrainMass!

28.

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.

24.

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:

F(a,b,c) = 16A^2 = p(p-2a)(p-2b)(p-2c), (1)

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

combination

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.