Answer:
- maximum: 0 on the line x=0
- minimum: -4 at (x, y) = (2, 0)
Step-by-step explanation:
You want the absolute extrema of f(x, y) = x²y² -2x over the triangular region bounded by (2, 0), (0, 2), (0, -2).
Extrema
The absolute extrema will lie at a critical point within the region, on the boundary, or at one of the corners of the region. The function values at the vertices of the region are ...
f(2, 0) = 2²·0² -2·2 = -4
f(0, ±2) = 0²·(±2)² -2·0 = 0
Critical points
The partial derivatives of the function are ...
[tex]f_x(x,y)=2xy^2-2\\\\f_y(x,y)=2x^2y[/tex]
A critical point will be found where these are zero simultaneously. For ∂f/∂y = 0, x = 0 or y = 0. For either of these values, ∂f/∂x is negative, so the function has no critical point.
General shape
We observe that the second derivative with respect to y is non-negative everywhere. This means the curve is concave upward.
∂²f/∂y² = 2x² . . . . . ≥ 0 on x ∈ [0, 2]
The first partial derivative is 0 for x=0 and y=0, so the line y=0 represents a minimum of the curve on any line x=c, 0 ≤ c ≤ 2.
Boundary lines
The lines bounding the triangle are x=0, and y=±(x -2). We already know that f(x, y) = 0 for x = 0. The values on the boundary lines will be ...
f(x, y) = f(x, ±(x -2)) = x²(±(x-2))² -2x = x⁴ -4x³ +4x² -2x
The derivative on this boundary line is ...
f'(x) = 4x³-12x² +8x -2
For x ∈ [0, 2], f'(x) is negative. This means there will be no boundary line points more extreme than those at the vertices of the region.
The extrema of the function are ...
- maximum: 0 on the line x=0
- minimum: -4 at (x, y) = (2, 0)
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Additional comment
The attachment is a graph of the function on the region of interest. This is consistent with the description above.
