Happy hour prices Part-2

We define a term $\phi(x) = \dfrac{\text{cost}}{\text{x litres}}$ .

$\phi(0.5) = \dfrac{3}{0.5} = 6 = \varphi_1 \\ \phi(1) = \dfrac{10}{1} = 10 = \varphi_2 \\ \phi(2) = \dfrac{20}{2} = 10 = \varphi_3$

We can see that $\varphi_1 < \varphi_2 = \varphi_3$ , We will have the maximum price when we use $\varphi_2 , \varphi_3$ to determine the maximum price.

We define $\Gamma_i=\varphi_i \delta \quad \forall i \ 1\leq i \leq 3$ where $\delta$ is the number of liters of drink ordered. We see that $\Gamma_i \propto \varphi_i$ and $\Gamma_i$ is maximum when $\varphi_i$ is maximum , since $\delta$ is fixed.So we have:

$\Large\Gamma_{i(max)} = \varphi_{i(max)}\delta = \varphi_{(2,3)}\delta = 10 \times 10 = \boxed{100}$

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Bonus question: Write an essay expounding the woes of a customer who did not learn mathematics.

If they ordered the $2$ liter $5$ times, they will get to drink $10$ liters fo $\$100$

Or, If they ordered the $1$ liter $10$ times, it won't differ.

There are many more ways to buy $10$ Liters for the condition in the question. Hint: Use permutations and combinations!

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Bonus question: Write an essay expounding the woes of a customer who did not learn mathematics.