Find the highest/lowest points on the ellipse which is the intersection of the p

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We want to find extrema of z subject to the constraints 4x - 3y + 8z = 5 and z^2 = x^2 + y^2. Since we have two constraints we will need two Lagrange multipliers. Let f(x, y, z) = z - ?_1 (4x - 3y + 8z - 5) - ?_2 (z^2 - x^2 - y^2). Then ?f/?x = -4?_1 + 2x?_2 = 0 ?f/?y = 3?_1 + 2y?_2 = 0 ?f/?z = 1 - 8?_1 - 2z?_2 = 0 Eliminating ?_1 from the first two equations gives us 6x?_2 + 8y?_2 = 0. Now it cannot be that ?_2 = 0, else we would have ?_1 = 0 from the first two equations and then the third equation would read 1 = 0. So we have 6x + 8y = 0, i.e. y = -3x/4. Eliminating ?_1 from the first and third equations gives us 4x?_2 + 2z?_2 - 1 = 0, or (4x + 2z)?_2 = 1. Eliminating ?_2 from the same equations gives us -4z?_1 + x - 8x?_1 = 0, or (8x + 4z)?_1 = x. So we have ?_1 = (x/2) ?_2, or x = 2?_1 / ?_2. We know y = -3/4 x, so y = -3?_1 / 2?_2. And from the third equation we have z = (1-8?_1) / 2?_2. Now we apply the constraints. 4x - 3y + 8z = 5 => (8?_1 / ?_2) + (9?_1 / 2?_2) + ((8 - 64?_1) / 2?_2) = 5 => (8 - 39?_1) / (2?_2) = 5 => 8 - 39?_1 = 10?_2. z^2 = x^2 + y^2 => (1 - 16?_1 + 64?^2_1) / (4?^2_2) = (16?^2_1 + 9?^2_1) / (4?^2_2) => 1 - 16?_1 + 64?^2_1 = 25?^2_1 => 39?^2_1 - 16?_1 + 1 = 0 => (13?_1 - 1) (3?_1 - 1) = 0 So ?_1 is 1/3 or 1/13. Case 1: ?_1 = 1/3 10?_2 = 8 - 39?_1 = 8 - 13 = -5, so ?_2 = -1/2. Then x = 2?_1 / ?_2 = -4/3; y = -3/4 x = 1; and 8z = 3y - 4x + 5 = 40/3, so z = 5/3. Case 2: ?_1 = 1/13 10?_2 = 8 - 39?_1 = 8 - 3 = 5, so ?_2 = 1/2. Then x = 2?_1 / ?_2 = 4/13; y = -3/4 x = -3/13; and 8z = 3y - 4x + 5 = 40/13, so z = 5/13. Hence the highest point on the ellipse is (-4/3, 1, 5/3) and the lowest point is (4/13, -3/13, 5/13).

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