course What does it mean to 'identify' a contour? The contour lines in the question you asked about are identified as circles with a given center and radius. Contour lines can be elliptical, hyperbolic, parabolic, etc., and you should be able to identify the ellipse, hyperbola, parabola, etc.. However if you get the equation and sketch the contour lines, you'll only lose a token point or two if you fail to identify the shapes (two points if it's a circle, since you should be able to identify the center and radius of a circle; probably just one if it's one of the other shapes). Dave, Thanks for the reply. It seems the correct approach is to find the partial derivatives, leading to the forms of the level curves being y=c x^-(3/2). But the same test problem I cited also states that the TOTAL COST of labor (at $125/unit) plus capital (at $100/unit) is limited to $100,000.00 How does this add to the complexity and how do we deal with this restriction? How do I calculate the final answer which is the max production level? That makes the problem much more interesting. With the information I apparently missed before, you know that 125 x + 100 y <= 100,000, or y <= -5/4 x + 1000. Geometrically this corresponds to the region of the first quadrant bounded by the line y = -5/4 x + 1000; the y intercept is 1000 and the x intercept is 800. You would want to find the level curve with the highest value of c that lies within this region. From the contour graph it appears that this would occur at the point of tangency between a level curve and the straight line. It would be possible to set up a system of three equations that could be solved simultaneously for this condition (conditions would be y = -5/4 x + 1000, y = c x^(-3/2) and the slope of the contour would be -5/4, so that the derivative of c x^(-3/2) would be -5/4; the third equation would therefore be -3/2 c x^(-5/2) = -5/4). The simultaneous solution of the equations wouldn't be overly difficult. Alternatively, since the geometry of the contours dictates a maximum on the straight line, we can use y = -5/4 x + 1000 to represent the function 200 x^.6 y^.4 along that line; we get g(x) = 200 x^.6 ( -5/4 x + 1000 )^.4. The derivative of this function is g'(x) = 120 x^(-.4) (-5/4 x + 1000) ^ .4 + 200 x^.6 (-5/4) * .4 (-5/4 x + 1000)^-.6. If we set g ' (x) = 0, then multiply both sides by the 'common denominator' x^.4 (-5/4 x + 1000)^.6 we get 120 (-5/4 x + 1000) - 100 (-5/4 x + 1000) = 0 or -150 x + 120 000 + 125 x + 100 000 = 0 so that 25 x = 20 000 and x = 400, so that y = -5/4 x + 1000 = 500. The maximum occurs at the point (x, y) = (400, 500). The value of the max would be obtained by just plugging these values into the original expression 200 x^.6 y^.4. I don't have time to check all the numbers, which were calculated mentally, but they do appear to be reasonably consistent. I do think the procedure given here is in any case valid.