Find the absolute extrema of the function on the closed triangular region with vertices , and .
A) Absolute minimum -5,Absolute maximum 5
B) Absolute minimum 0,Absolute maximum 5
C) Absolute minimum -5,Absolute maximum 17
D) Absolute minimum 5,Absolute maximum 17

Question

Find the absolute extrema of the function  on the closed triangular region with vertices  ,  and  .
A) Absolute minimum -5,Absolute maximum 5 
B) Absolute minimum 0,Absolute maximum 5 
C) Absolute minimum -5,Absolute maximum 17 
D) Absolute minimum 5,Absolute maximum 17

Quizplus · Accepted Answer

Let the vertices of the triangle be denoted by $A$, $B$, and $C$, respectively. Using the formula for the distance between two points in the coordinate plane, we can find that the side lengths of the triangle are: $AB = \sqrt{2^2 + 3^2} = \sqrt{13}$, $AC = \sqrt{(-1)^2 + 4^2} = \sqrt{17}$, and $BC = \sqrt{(1)^2 + (1)^2} = \sqrt{2}$. So, the base of the triangle is $BC = \sqrt{2}$, and the height is given by the equation $y = -\frac{\sqrt{3}}{2}x + 3\sqrt{3}$. We can find the maximum and minimum values of the function over the triangle by checking the values of the function at the vertices and at critical points inside the triangle. At the vertices: $f(A) = f(-2,3) = -2 < f(1,1) = 5 < f(-1,4) = 17$ So the maximum value of $f$ over the triangle is 17. $f(A) = -2 > f(1,1) = 5 > f(-1,4) = 17$ So the minimum value of $f$ over the triangle is -2. Therefore, the answer is (C), with the absolute minimum value of $-5$ an incorrect choice, and the absolute maximum value of $5$ an incorrect choice.

Find the Absolute Extrema of the Function on the Closed