Question

find the absolute minimum and absolute maximum of

$f(x,y)=18 - 5x + 8y$

on the closed triangular region with vertices $(0,0),(8,0)$ and $(8,13)$.

list the minimum/maximum values as well as the point(s) at which they occur. if a min or max occurs at multiple points separate the points with commas.

minimum value:

occurs at

maximum value:

occurs at

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Explanation:

Step1: Analyze the function on the boundaries

• Boundary 1: \(y = 0\), \(0\leq x\leq8\)

The function becomes \(f(x,0)=18 - 5x\). Since the derivative \(f^\prime(x,0)=-5<0\), it is a decreasing function. At \(x = 8\), \(f(8,0)=18-5\times8=18 - 40=-22\); at \(x = 0\), \(f(0,0)=18\).

• Boundary 2: \(x = 8\), \(0\leq y\leq13\)

The function becomes \(f(8,y)=18-5\times8 + 8y=8y - 22\). Since the derivative \(f^\prime(8,y)=8>0\), it is an increasing function. At \(y = 0\), \(f(8,0)=-22\); at \(y = 13\), \(f(8,13)=8\times13-22=104 - 22 = 82\).

• Boundary 3: Find the equation of the line passing through \((0,0)\) and \((8,13)\)

The slope \(m=\frac{13-0}{8 - 0}=\frac{13}{8}\), and the equation is \(y=\frac{13}{8}x\). Substitute \(y=\frac{13}{8}x\) into \(f(x,y)\): \(f(x,\frac{13}{8}x)=18-5x + 8\times\frac{13}{8}x=18-5x+13x=18 + 8x\). Since the derivative \(f^\prime(x,\frac{13}{8}x)=8>0\), it is an increasing function. At \(x = 0\), \(f(0,0)=18\); at \(x = 8\), \(f(8,13)=82\).

Step2: Check the critical points

Find the partial derivatives: \(\frac{\partial f}{\partial x}=-5\), \(\frac{\partial f}{\partial y}=8\). Since \(\frac{\partial f}{\partial x}
eq0\) and \(\frac{\partial f}{\partial y}
eq0\) simultaneously, there are no critical points in the interior of the triangular region.

Answer:

Minimum value: \(-22\)
Occurs at \((8,0)\)
Maximum value: \(82\)
Occurs at \((8,13)\)