2) To write the constraint equation, think about the restriction put on the variables. We cannot choose any two numbers for x and y. Since Ayesha has 32 ft of fencing, the distance around the garden is 32 ft. This is the perimeter of the rectangular garden. The perimeter of the rectangle drawn above is 2x 2y, and it must equal 32 ft. The constraint equation is Constraint: 2x 2y 32 Set up this maximization problem as Maximize: A xy Constraint: 2x 2y 32 Solve the constraint for a variable, and then substitute the expression into the maximize equation. 2x 2y 32 2y 32 2x y 16 x Solve the constraint for y. Substitute y 16 x into A xy. A x(16 x) A 16x x2 Distribute. A x2 16x Write in descending powers. Look carefully at A x2 16x. This is a quadratic function! Its graph is a parabola that opens downward (since a 1). At the vertex, the function attains its maximum. The ordered pairs that satisfy this function are of the form (x, A(x)), where x represents the width and A(x) represents the area of the rectangular garden. The second coordinate of the vertex is the maximum area we are looking for. A x2 16x Use x b 2a with a 1 and b 16 to fi nd the x-coordinate of the vertex (the width of the rectangle that produces the maximum area). x 16 2(1) 8 Substitute x 8 into A x2 16x to fi nd the maximum area. A (8)2 16(8) A 64 128 A 64 The graph of A x2 16x is a parabola that opens downward with vertex (8, 64). The maximum area of the garden is 64 ft2, and this will occur when the width of the garden is 8 ft. (The length will be 8 ft as well.) Let’s summarize the steps we can use to solve a max/min problem. Do you understand how the solution of this problem is related to the graph of a quadratic equation? SECTION 10.6 Applications of Qu www.mhhe.com/messersmith adratic Functions and Graphing Other Parabolas 671
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