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 with parallel to the partition be y metres.

Based on this information, answer the following questions:

1. Write the equation for the total boundary material used in the boundary and parallel to the partition in terms of x and y.   [1]
2. Write the area of the solar panel as a function of x.   [1]
3. 1. Find the critical points of the area function. Use second derivative test to determine critical points at the maximum area. Also, find the maximum area.   [2]
                                                         OR
   2. Using first derivative test, calculate the maximum area the company can enclose with the 300 metres of boundary material, considering the parallel partition.   [2]

Answer 1

(i)

Given, perimeter = 300 metres

x + x + y + y + y = 300   

2x + 3y = 300      ...(i)

(ii) Area of solar panel (A) = 1 × b 

A = x × y

A = xy       ...(ii)

From equation (i),

3y = 300 − 2x

y = `(300 - 2x)/3`

Put the value y in equation (ii),

∴ A = `x((300 - 2x)/3) m^2`

(iii) (a) A = `(300x - 2x^2)/3`

`(dA)/(dx) = 1/3(300 - 4x)`

`(d^2A)/(dx^2) = (-4)/3 < 0 ("maximum")`

For critical point

`(dA)/(dx) = 0`

`1/3 (300 - 4x) = 0`

300 = 4x

Then, x = `300/4`

= 75 m

Maximum area = `1/3[300 xx 75 - 2(75)^2]`

= `1/3[22500 - 2 xx 5625]`

= `1/3[22500 - 11250]`

= `11250/3`

= 3750 m²

OR

(iii) (b) For critical point

`(dA)/(dx) = 0`

`1/3(300 - 4x) = 0`

4x = 300

x = `300/4`

x = 75

By using first order derivative

`A'(74) = 1/3(300 - 4 xx 74)`

= `1/3(300 - 296)`

= `4/3 > 0`

`A'(76) = 1/3(300 - 4 xx 76)`

= `1/3 (300 - 304)`

= `(-4)/3 < 0`

Derivative changes from +ve to – ve at 75.

∴ Area is maximum at x = 75.