<p>A rectangular area is to be fenced against an existing wall. The three sides of the fence must be a total of 1050 feet long. Find the dimensions of the maximum area that can be enclosed. What is the maximum area?</p>
<p>2)</p>
<p>The ventilating system for the large group room requires ducts with rectangular cross sections. These are to be made by folding sheet-metal strips that are 62 in. wide. The volume of air in the ducts can be maximized by designing the ducts to have maximum cross-sectional area. What are the best dimensions for these ducts? What is the maximum cross-sectional area?</p>
<p>1) Let x be the length of the side perpendicular to the wall.<br>
The area then is x(1050-2x) = 1050x - 2x^2.</p>
<p>At this point, you have at least three options to find the maximum.</p>
<p>First, you could set the derivative 1050-4x to zero (assuming you know calculus).</p>
<p>Second, you could plug the expression into your graphing calculator.</p>
<p>Third, you could recognize the expression as describing a concave down parabola with solutions for X= and X= 525. From symmetry, you would place the vertex at X = 262.5</p>
<p>In any event, plug X = 262.5 into your area expression to get maximum area 137,550 square feet.</p>
<p>First, you might know that the area of an arbitrary rectangle of fixed perimeter is maximized when the sides are equal. This requires that the side length be 62/4 inches with resultant cross sectional area of 240.25 square inches.</p>
<p>Second, let x = the dimension of one side of the rectangle; the cross sectional area then is equal to x((62-2x)/2) = x(31-x).</p>
<p>We have the same three methods for finding the maximum area available:
Differentiate the area expression, find a zero for x = 15.5; plug into calculator; recognize expression as parabola with zeros at x = 0 and x = 31 (so vertex is at x=15.5)</p>
<p>side = 15.5, so max area = 240.25 square inches</p>