If $p$ and $q$ vary such that $3p + 2q = 50$, find the maximum value of $10pq$

Question

Let $p$ and $q$ be real numbers such that $3p+2q=50$.

Find a maximal value of $10pq$.

I don't quite get this question. This is from a high school math exam paper, so partial derivatives are out. I initially thought of squaring the whole equation, but I realized I end up with $9p^2$ and $4q^2$, which I cannot eliminate.

Any help will be appreciated.

Note: I'm also unsure of the topic this falls into, if someone can verify for me that'll be great.

Answer 1

It $3p+2q=50$, then $q=25-\frac32 p$ meaning that $$10pq=10p(25-\frac32 p) = 250p - 15p^2$$

Now all you need to do is to find the maximum of a single-variable quadratic function, something a high school student should be capable of.

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Remember, if you know the zeroes of a quadratic function, then the maximum or minimum lies directly between the two. The zeroes of $$10p(25-\frac32 p) = 5p(50-3p)$$

are $p=0$ and $p=\frac{50}{3}$.

Answer 2

We can use $xy\leq\left(\frac{x+y}{2}\right)^2$.

$$10pq=\frac{5}{3}\cdot3p\cdot2q\leq\frac{5}{3}\left(\frac{3p+2q}{2}\right)^2=\frac{5^5}{3}.$$ The equality occurs for $3p=2q$ and $3p+2q=50$, which says that $\frac{3125}{3}$ is a maximal value.

Answer 3

A solution to $3p+2q=50$ is $(p,q) = (0,25)$.

So a general solution would be $p = 0+2t$ and $q = 25 - 3t$.

Then $10pq = 20t(25-3t)$; a parabola whose vertex is half way between $t=0$ and $t=\frac{25}{3}$.

Setting $t=\frac{25}{6}$, we get that the max value is $\frac{3125}{3}$