Answered Essay: In the rectangle problem, a feasible alternative must satisfy the condition w + h = with wand 

sider forming a maximum-area rectangle out of a piece of wire of length L inches. What should be the width and height of the rectangle? In contrast with the tickets example, the number of alternatives in the present ex- ample is not finite; namely, the width and height of the rectangle can assume an infinite number of values. To fornalize this observation, the alternatives of the problem are identified by defining the width and height as continuous (algebraic) variables. Let w-width of the rectangle in inches h = height of the rectangle in inches Based on these definitions, the restrictions of the situation can be expressed verbally as 1. Width of rectangle + Height of rectangle - Half the length of the wire 2. Width and height cannot be negative These restrictions are translated algebraically as

In the rectangle problem, a feasible alternative must satisfy the condition w + h = with wand h assuming nonnegative values. This leads to an infinite number of feasible solutions and, unlike

the tickets problem, the optimum solution is determined by an appropriate mathemat- ical tool (in this case, differential calculus).

Question: In the rectangle problem, identify two feasible solutions and determine which one is better?

Sider forming a maximum-area rectangle out of a piece of wire of length L inches. What should be the width and height of the rectangle? In contrast with the tickets example, the number of alternatives in the present ex- ample is not finite; namely, the width and height of the rectangle can assume an infinite number of values. To fornalize this observation, the alternatives of the problem are identified by defining the width and height as continuous (algebraic) variables. Let w-width of the rectangle in inches h = height of the rectangle in inches Based on these definitions, the restrictions of the situation can be expressed verbally as 1. Width of rectangle + Height of rectangle – Half the length of the wire 2. Width and height cannot be negative These restrictions are translated algebraically as

Expert Answer

Answer

Considering the constraints : w+h = L/2 and w>0 ,h>0

and assuming that the wire of length L is forming a rectangle which means two adjacent sides are of different length

Hence two feasible solutions would be w= L/8 ,h=3L/8 and w=L/6 ,h=2L/6 Both of these satisfy the above constraints.

Now our aim is to maximize the area hence case 1 : w= L/8 ,h=3L/8 , Area =w*h = 3/64 *L2 = 0.046875 L2

Case 2 : w=L/6 ,h=2L/6 ,Area = w*h = 2/36 * L2 = 0.0555 L2

Now since O.0555>0.046875 Area is more for case 2 hence case 2 is better considering our aim

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