Consider the girth first, it is the shortest distance around the parcel. And then L is the longest side. Based on the dimensions, this gives us two sets of possible equations. The first, if x is the longest side, gives us this:
Or, if y is the longest:
Alright, I left off here! Let me know if you need a hand with the other question, it is just differentiation and finding a maxima, give it a go!
So here, we have two potential equations to start with, let's see how they effect the formulae for volume (in both cases, we'll eliminate y):
When we do this, it becomes a little clearer that the top option gives a larger volume for any value of x. I'll leave you to form a mathematical proof with all this information, but that's the basic setup!
To show you the idea of finding the maxima/minima, I'll work through this one for you:
=0 \\\therefore x=0,\frac{50}{3} \\ x\neq0\quad \therefore x=\frac{50}{3} \\ x=\frac{50}{3} \implies \frac{d^2V}{dx^2}=200-400<0 \\\therefore\text{ volume is maximised at }x=\frac{50}{3} )
So the dimensions are (using the formulas above), 50/3 by 50/3 by 100/3
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