2024 DHS H1 Q5

[It is given that a sphere of radius \(r\) has surface area \(4\pi r^2\) and volume \(\frac{4}{3}\pi r^3\).]

A manufacturer builds storage containers for AgriSupply, an agricultural supply company that sells liquid fertilizer. Each container has a volume of \(480\) \(\mathrm{m}^3\). The container consists of a cylindrical body with length \(h\) \(\mathrm{m}\) and two identical hemispherical caps at the two ends, each with radius \(r\) \(\mathrm{m}\), as shown in the diagram above.

1. Show that the total surface area, \(A\) \(\mathrm{m}^2\), of the container is given by\[A = \frac{960}{r} + \frac{4}{3}\pi r^2.\][3]
2. Use differentiation to find the minimum value of \(A\), giving your answer to \(2\) decimal places. Justify that this is a minimum value.[4]
3. AgriSupply sells three specialized fertilizer blends: Fertimax \(X\), Fertimax \(Y\) and Fertimax \(Z\). Each blend is formulated to meet specific nutrient requirements for different stages of crop growth. A farm manager is given the following information about their prices.

For the minimisation, include the limiting design with cylindrical height zero, where the two hemispheres form a sphere. If the cylindrical height must be strictly positive, the same surface-area bound is an infimum and is not attained.

1. The total cost for \(40\) cans of Fertimax \(X\), \(15\) cans of Fertimax \(Y\), and \(16\) cans of Fertimax \(Z\) is \(\$25250\).
2. The cost of \(8\) cans of Fertimax \(Z\) is the cost of \(5\) cans of Fertimax \(X\).
3. The total cost for \(20\) cans of Fertimax \(Y\) is \(\$3200\) more than the combined cost of \(7\) cans of Fertimax \(X\) and \(4\) cans of Fertimax \(Z\).

Determine the greatest number of cans of Fertimax \(Y\) that can be purchased with a budget of \(\$10000\).[5]