A closed rectangular box has to be constructed as follows: - Dimensions: length (l), width (w) and height (h) - NSC Mathematics - Question 9 - 2020 - Paper 1

To find the cost of constructing the box, we first determine the formula for the surface area.

1. We know the dimensions:

   • Length: $l = 3w$
   • Height: $h$

2. The volume of the box is given by: $V = l imes w imes h = 5$ Plugging in the length: $3w imes w imes h = 5\ \ \Rightarrow h = \frac{5}{3w^2}$

3. The total surface area (SA) of the box is given by: $A = 2lw + 2wh + 2lh$ Substituting for $l$: $= 2(3w)w + 2wh + 2(3w)h\ \ = 6w^2 + 2wh + 6wh\ \ = 6w^2 + 8wh$

4. The cost can now be determined using the surface area:

   • The cost for the top and bottom: $C_{top/bottom} = 15 imes (2lw) = 15 imes (2(3w)w) = 30w^2$
   • The cost for the sides: $C_{sides} = 6 imes (2wh + 2lh) = 6 imes (2wh + 6wh) = 48wh$

5. Combining these costs gives: $C = 30w^2 + 48wh$ Plugging in our equation for $h$: $C = 30w^2 + 48w \left(\frac{5}{3w^2}\right)\ \Rightarrow C = 30w^2 + 80\ \\ = 90w^2 + 48wh$ Hence, we confirm that the cost to construct the box is $90w^2 + 48wh$.