Question

7. (a) (2 points) Sketch the level curves of the function f(x,y) 4-2 - 2y2, at levels k 4,3, 0, and-5. (b) (3 points) What region R in the ry-plane maximizes the value of (4 - 2 -2y2) dA? Give reasons for your answer (c) (5 points) Then, express the double integral over the region R you specified above as an iterated integral (no need to compute it)

I need help with this question and some concept clarifications would be great.

Answer 1

4 = 4 - x^2 - 2y^2 doublerightarrow -x^2 - 2y^2 = 0 doublerightarrow x^2 + 2y^2 = 0 so 4 = 4 - x^2 - 2y^2 doublerightarrow (x, y) = (0, 0) 3 = 4 - x^2 - 2y^2 -1 = -x^2 - 2y^2 doublerightarrow x^2 2y^2 = 1 doublerightarrow x^2/1^2 + y^2/(y/squareroot 2)^2 = 1 (Ellipse with major axis 1 minor axis 1/squareroot 2) 0 = 4 - x^2 - 2y^2 doublerightarrow x^2 + 2y^2 = 4 doublerightarrow x^2/2^2 + y62/(squareroot 2)^2 = 1 (Ellipse with major axis 2 minor axis squareroot 2) -5 = 4 - x^2 - 2y^2 -9 = -x^2 - 2y^2 x^2 + 2y^2 = 9 x^2/9 + 2y^2/9 = 1 x^2/3^2 + y^2/(3/squareroot 2)^2 = 1 (Ellipse with major axis 3 minor axis 3/squareroot 2) level curves at different k = 4, 3, 0, -5 \r\n doubleintegral_R (4 - x^2 - 2y^2) dA is maximized in region R in which 4 - x^2 - 2y^2 greaterthanorequalto 0 because if 4 - x^2 - 2y^2 < 0 then integration will decrease \r\n < null >