you have 200 feet of fencing to enclose a rectangular plot that borders on a river. if you do not fence the…

you have 200 feet of fencing to enclose a rectangular plot that borders on a river. if you do not fence the side along the river, find the length and width of the plot that will maximize the area. what is the largest area that can be enclosed?\nthe width, labeled x in the figure, is 50 feet.\n(type an integer or decimal.)\nthe length, labeled 200 - 2x in the figure, is 100 feet.\n(type an integer or decimal.)\nthe largest area that can be enclosed is square feet.\n(type an integer or decimal.)

you have 200 feet of fencing to enclose a rectangular plot that borders on a river. if you do not fence the side along the river, find the length and width of the plot that will maximize the area. what is the largest area that can be enclosed?\nthe width, labeled x in the figure, is 50 feet.\n(type an integer or decimal.)\nthe length, labeled 200 - 2x in the figure, is 100 feet.\n(type an integer or decimal.)\nthe largest area that can be enclosed is square feet.\n(type an integer or decimal.)

Answer

Explanation:

Step1: Write the area formula

The area (A) of a rectangle is (A = length\times width). Given (width=x) and (length = 200 - 2x), then (A(x)=x(200 - 2x)=200x-2x^{2}).

Step2: Find the vertex of the quadratic function

For a quadratic function (y = ax^{2}+bx + c) ((a=- 2), (b = 200), (c = 0)), the (x) - coordinate of the vertex is (x=-\frac{b}{2a}). Substitute (a=-2) and (b = 200) into (x =-\frac{b}{2a}), we get (x=-\frac{200}{2\times(-2)} = 50). We already know (x = 50) (width), (length=200-2x). Substitute (x = 50) into (length) formula: (length=200-2\times50=100). Now substitute (x = 50) into the area formula (A(x)=200x-2x^{2}). (A(50)=200\times50-2\times50^{2}) (=10000-2\times2500) (=10000 - 5000)

Answer:

The largest area that can be enclosed is (5000) square feet.