Question

a lifeguard needs to rope off a rectangular swimming area in front of long lake beach, using 1200 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)=1200x - 2x^{2} ) (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

Explanation:

Step1: Analyze the objective function

The area function is \(A(x)=1200x - 2x^{2}\), which is a quadratic function of the form \(y = ax^{2}+bx + c\) (\(a=- 2\), \(b = 1200\), \(c = 0\)). For a quadratic function \(y=ax^{2}+bx + c\), the vertex of the parabola (which gives the maximum/minimum value when \(a
eq0\)) has its \(x\) - coordinate at \(x=-\frac{b}{2a}\).

Step2: Find the value of \(x\) (shorter side)

Substitute \(a=-2\) and \(b = 1200\) into the formula \(x=-\frac{b}{2a}\).

$$x=-\frac{1200}{2\times(-2)}=\frac{1200}{4}=300$$

Step3: Find the length of the longer side

Let the length of the longer side be \(y\). We know that the total length of the rope is \(1200\) yd. If the two sides perpendicular to the shoreline have length \(x\) each, then \(y + 2x=1200\). Substitute \(x = 300\) into the equation \(y=1200 - 2x\).

$$y=1200-2\times300=600$$

Answer:

The length of the shorter side of the rectangular region is \(300\) yd.
The length of the longer side of the rectangular region is \(600\) yd.