A farmer wishes to enclose a pasture that is bordered on one side by a river(only 3 sides require fencing), She has decided to make a rectangular shape and will use barbed wire to make it. There are 600 ft of wire for the project, and will use all the wire .What is the maximum area that can be enclosed by the fence?(hint:use info to create a quadratic function for the area then find the maximum of the function).
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sum of all three border : 2x+y = 600 hence we can write , y = 600-2x
Area of rectangular land, A = xy
or we can write A = x(600-2x) ; A = 600 x - 2x²
To find the maximum possible area, let us find its first derivative
A' = 600-4x = 0
solving we get , 4x = 600 ; or x = 150 ft y = 600- 2x = 300 ft
Maximum possible Area = 150×300 = 45000 square ft
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Since A represents a quadratic equation hence we can re-write A the exponents in descending order:
A =- 2x² +600 x
The graph of A will be parabola and, since , the parabola will have a maximum point as its vertex. The y-coordinate of the vertex will represent our greatest area. To proceed, we need to find the value of the x-coordinate of the vertex (that is, the value of l in our equation).