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.
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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.
(i) The total boundary material comprises the perimeter of a rectangular solar panel plus an internal partition parallel to one side. The perimeter is \( 2x + 2y \), and the partition length is \( y \). Therefore, the total boundary material is \( 2x + 2y + y = 2x + 3y \). Given 300 metres of material, the equation is \( 2x + 3y = 300 \).
(ii) The solar panel's area \( A \) is \( x \times y \). From the boundary equation \( 2x + 3y = 300 \), we get \( y = \frac{300 - 2x}{3} \). Substituting this into the area equation yields \( A(x) = x \times \frac{300 - 2x}{3} = \frac{300x - 2x^2}{3} \). The area as a function of \( x \) is \( A(x) = \frac{300x - 2x^2}{3} \).
(iii) To find critical points, differentiate \( A(x) \) with respect to \( x \): \( A'(x) = \frac{d}{dx} \left( \frac{300x - 2x^2}{3} \right) = \frac{1}{3}(300 - 4x) \). Setting \( A'(x) = 0 \) gives \( \frac{300 - 4x}{3} = 0 \), so \( 300 - 4x = 0 \), which means \( x = 75 \). The second derivative is \( A''(x) = \frac{d}{dx} \left( \frac{1}{3}(300 - 4x) \right) = \frac{-4}{3} \). Since \( A''(x)<0 \), \( x = 75 \) corresponds to a maximum.
Substitute \( x = 75 \) into \( 2x + 3y = 300 \): \( 2(75) + 3y = 300 \implies 150 + 3y = 300 \implies 3y = 150 \implies y = 50 \). The maximum area is \( A = 75 \times 50 = 3750 \) square metres.
OR
(iii) Using the first derivative test, \( A'(x) = \frac{300 - 4x}{3} \). For \( x<75 \), \( A'(x)>0 \) (area increasing). For \( x>75 \), \( A'(x)<0 \) (area decreasing). Thus, \( x = 75 \) yields a maximum area of 3750 square metres, as previously calculated.
