Question

A technical company is designing a rectangular solar panel installation on a roof using 300 metres of boundary material. The design includes a partition running parallel to one of the sides dividing the area (roof) into two sections.
Let the length of the side perpendicular to the partition be \( x \) metres and the side parallel to the partition be \( y \) metres.
Based on this information, answer the following questions:
(i) Write the equation for the total boundary material used in the boundary and parallel to the partition in terms of \( x \) and \( y \).
(ii) Write the area of the solar panel as a function of \( x \).
(iii) Find the critical points of the area function. Use the second derivative test to determine critical points at the maximum area. Also, find the maximum area.
OR
(iii) Using the first derivative test, calculate the maximum area the company can enclose with the 300 metres of boundary material, considering the parallel partition.

Hint:

To find the maximum area of a rectangular shape with a fixed boundary material, use the first and second derivative tests. Solving for \( y \) in terms of \( x \) from the perimeter equation is key.

Answer 1

(i) The total boundary material used includes the perimeter of the rectangular solar panel with an additional partition running parallel to one of the sides. The perimeter of the rectangle is \( 2x + 2y \), and the length of the partition is \( y \). Therefore, the total boundary material used is: \[ \text{Total Boundary Material} = 2x + 2y + y = 2x + 3y. \] We are given that the total boundary material used is 300 metres, so: \[ 2x + 3y = 300. \] (ii) The area \( A \) of the solar panel is the product of its length and width: \[ A = x \times y. \] From the boundary equation \( 2x + 3y = 300 \), solve for \( y \) in terms of \( x \): \[ y = \frac{300 - 2x}{3}. \] Substitute this expression for \( y \) into the area equation: \[ A(x) = x \times \frac{300 - 2x}{3} = \frac{300x - 2x^2}{3}. \] Thus, the area of the solar panel as a function of \( x \) is: \[ A(x) = \frac{300x - 2x^2}{3}. \] (iii) To find the critical points of the area function, we first differentiate \( A(x) \) with respect to \( x \): \[ A'(x) = \frac{d}{dx} \left( \frac{300x - 2x^2}{3} \right) = \frac{1}{3} \times (300 - 4x). \] Set \( A'(x) = 0 \) to find the critical points: \[ \frac{300 - 4x}{3} = 0 \implies 300 - 4x = 0 \implies x = 75. \] Now, we check the second derivative \( A''(x) \) to determine whether this critical point is a maximum or minimum: \[ A''(x) = \frac{d}{dx} \left( \frac{1}{3} \times (300 - 4x) \right) = \frac{-4}{3}. \] Since \( A''(x)<0 \), the critical point \( x = 75 \) corresponds to a maximum. Substitute \( x = 75 \) into the equation \( 2x + 3y = 300 \) to find \( y \): \[ 2(75) + 3y = 300 \implies 150 + 3y = 300 \implies 3y = 150 \implies y = 50. \] Thus, the maximum area is: \[ A = x \times y = 75 \times 50 = 3750 \, \text{square metres}. \] OR
(iii) To calculate the maximum area using the first derivative test, we observe that the first derivative is: \[ A'(x) = \frac{300 - 4x}{3}. \] For \( x<75 \), \( A'(x)>0 \), indicating that the area is increasing. For \( x>75 \), \( A'(x)<0 \), indicating that the area is decreasing. Therefore, \( x = 75 \) is a maximum, and the maximum area is 3750 square metres, as calculated earlier.