The absolute minimum of $f ( x , y ) = 2 x ^ { 2 } + y ^ { 2 } - 4 x - 4 y + 1$ on or inside the triangle with vertices (0,0), (1,2), and (0,2) is

A)1137
B)-61
C)-5
D)4
E)1

Question

The absolute minimum of $f ( x , y ) = 2 x ^ { 2 } + y ^ { 2 } - 4 x - 4 y + 1$ on or inside the triangle with vertices (0,0), (1,2), and (0,2) is

A)1137
B)-61
C)-5
D)4
E)1

Quizplus · Accepted Answer

The function $f(x, y) = 2x^2 + y^2 - 4x - 4y + 1$ is a quadratic function in two variables, which represents a paraboloid. To find its absolute minimum on or inside a given triangle, we first find the critical point by setting the partial derivatives equal to zero:$$\frac{\partial f}{\partial x} = 4x - 4 = 0 \implies x = 1$$$$\frac{\partial f}{\partial y} = 2y - 4 = 0 \implies y = 2$$The critical point is (1, 2), which lies inside the triangle. Evaluating $f$ at this point gives:$$f(1, 2) = 2(1)^2 + (2)^2 - 4(1) - 4(2) + 1 = 2 + 4 - 4 - 8 + 1 = -5$$Since this is a quadratic function (and thus convex), and the critical point is inside the triangle, the minimum value of $f$ on or inside the triangle is $f(1, 2) = -5$.

The Absolute Minimum Of f(x,y)=2x2+y2−4x−4y+1f ( x , y ) = 2 x ^ { 2 } + y ^ { 2 } - 4 x - 4 y + 1f(x,y)=2x2+y2−4x−4y+1

The Absolute Minimum Of $f ( x , y ) = 2 x ^ { 2 } + y ^ { 2 } - 4 x - 4 y + 1$