a lifeguard needs to rope off a rectangular swimming area in front of long lake beach, using 1800 yd of rope…

a lifeguard needs to rope off a rectangular swimming area in front of long lake beach, using 1800 yd of rope and floats. what dimensions of the rectangle will maximize the area? what is the maximum area? (note that the shoreline is one side of the rectangle.) let x be the length of a side of the rectangle perpendicular to the shoreline. write the objective function for the area in terms of x a(x)= (type an expression using x as the variable.) the length of the shorter side of the rectangular region is the length of the longer side of the rectangular region is the maximum area of the rectangular region is

a lifeguard needs to rope off a rectangular swimming area in front of long lake beach, using 1800 yd of rope and floats. what dimensions of the rectangle will maximize the area? what is the maximum area? (note that the shoreline is one side of the rectangle.) let x be the length of a side of the rectangle perpendicular to the shoreline. write the objective function for the area in terms of x a(x)= (type an expression using x as the variable.) the length of the shorter side of the rectangular region is the length of the longer side of the rectangular region is the maximum area of the rectangular region is

Answer

Explanation:

Step1: Express the length of the side parallel to the shoreline

Let ( x ) be the length of a side perpendicular to the shoreline. The total length of the rope is ( 1800) yd. Since there are two sides perpendicular to the shoreline ((x) each) and one side parallel to the shoreline. Let ( y ) be the length of the side parallel to the shoreline. Then ( 2x + y=1800), so ( y = 1800 - 2x).

Step2: Write the area function

The area of a rectangle (A=xy). Substitute (y = 1800 - 2x) into the area formula. So (A(x)=x(1800 - 2x)=1800x-2x^{2}).

Step3: Find the vertex of the quadratic function

For a quadratic function (y = ax^{2}+bx + c) ((a=- 2), (b = 1800), (c = 0)), the (x) - coordinate of the vertex is given by (x=-\frac{b}{2a}). [x=-\frac{1800}{2\times(-2)}=\frac{1800}{4} = 450]

Step4: Find the length of the side parallel to the shoreline

Substitute (x = 450) into (y=1800 - 2x). Then (y=1800-2\times450=1800 - 900=900)

Step5: Find the maximum area

Substitute (x = 450) into (A(x)=1800x-2x^{2}). [A(450)=1800\times450-2\times(450)^{2}] [=810000-2\times202500] [=810000 - 405000=405000]

Answer:

The objective function (A(x)=1800x - 2x^{2}). The length of the shorter side (perpendicular to the shoreline) is (450) yd. The length of the longer side (parallel to the shoreline) is (900) yd. The maximum area is (405000) square - yd.