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Cost%20Curve%2C%20min%20avg%20cost%2C%20and%20slope%20of%20tangent%20line

# Cost%20Curve%2C%20min%20avg%20cost%2C%20and%20slope%20of%20tangent%20line

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Unformatted text preview: The curve above is your cost func2on –> C(q). Please recall that it is the TOTAL cost of q items produced. The x ­axis represents “q” the number of items produced, while “C”, the cost of epresented t hose i tems i s r by t he y ­ a xis. The coordinates of any point on that line are therefore ( q,C (q)). To ﬁnd a(q), the average cost at any 2me, we merely divide that total cost C(q) by the number of items produced. The basic equa2on is a ( q ) = C qq ) . So let’s ﬁnd the ( slope o f a l ine c b y d rawing t he l ine b etween a ny p o n t he c urve and t origin. reated oint he The slope of that line is given below. Note that this is equal to a(q). m= C (q ) ! 0 C (q ) = q!0 q (Press any key or click mouse to con2nue) (q,C) Now let’s add a few more lines, each of which intersects our Cost func2on at a single point. NOTE: The tangent line has the SMALLEST slope. Recall that a horizontal line has a slope of 0. The closer a line is to horizontal the smaller the slope. Also recall that the slope of this line is equal to the average cost. Therefore the coordinates of the point where the line is tangent to the curve represent the SMALLEST slope of any line through any point on the curve. o p , oint Therefore t he c oordinates f t hat oint r epresented b y t he p ( q,C ) a bove, give the lowest average cost. This minimum average cost occurs when producing “q” items, where q is the coordinate of the point at the tangent line. One ﬁnal note, the tangent line must NOT touch the cost func2on at any other point for this demonstra2on to be valid. (Press any key or click mouse to con2nue) ...
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