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PROBLEM 16 : What angle between two edges of length 3 will result in an isosceles triangle with the largest area ? (See diagram.) SOLUTION Write the area of the given isosceles triangle as a function of . Let variable x be the length of the base and variable y the height of the triangle, and consider angle . Write each of x and y as functions of . It follows from basic trigonometry that so that (Equation 1 ) , and so that (Equation 2 ) We wish to MAXIMIZE the AREA of the isosceles triangle A = (1/2) (length of base) (height) = (1/2) xy . Before we differentiate, use Equations 1 and 2 to rewrite the right-hand side as a function of only. Then A = (1/2) xy . Now differentiate this equation using the product rule and chain rule, getting (Factor out (9/2) and simplify the expression.) =0, so that and . It follows algebraically (Why ?) that so that from basic trigonometry we get or , or . and hence Because measures an angle in a triangle, it is logical to assume that . Thus, . See the adjoining sign chart for A' . If radians = 90 degrees, then A = 9/2 is the largest possible area for the triangle.