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Question'answers seminar 1 Answers 9/24/2010 Bus 14B – Question and Answer Seminar #1 (Fall ‘10) A1) Maximizing Word Problem – Solve by Algebra Find the dimensions of a rectangular field of maximum area that can be made from 480 m of fencing material. (See 14.2) Given: Perimeter so 2x + 2y = 480 Goal : Maximize area of field Area = x * y A(x) = x * y A(x) = x ( 240 – x) A(x) = -x 2 + 240x This is a parabola opening downward. Highest or max point is the vertex point. X value at the vertex point Is the axis of symmetry. X = -B/(2A) X = -240/-2 = 120 m. To find y, substitute into perimeter equation. Y = 120m. Dimensions are 120 x 120. A2) To Solve for Maximum or Minimum by Calculus a) Take the Derivative of the function. b) Set derivative equal to zero. c) Solve for “x”. Derivative refers to the slope of curve, instantaneous rate of change, slope of tangent line at point on the curve. Symbols for the derivative: y’, f’(x), x y The formula below is used to find the derivative. But in Ch. 12, you are given many rules. The first one, power rule, is shown in the next section. It is much easier to use the rules, than do use

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