MHB Finding Side Lengths: Where to Begin?

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To find the side lengths of a cuboid given its volume, the height \( h \) is defined, leading to the volume equation \( V = 100xyh \). The relationships for width \( w \) and length \( \ell \) are established as \( w = \frac{3y}{h} \) and \( \ell = \frac{\frac{x}{3}}{h} \). By multiplying these equations, it is shown that \( \ell w = \frac{xy}{h^2} \). Substituting back into the volume equation results in \( h^2 = \frac{1}{100} \), giving \( h = \frac{1}{10} \) and simplifying the volume to \( V = 10xy \). This process effectively demonstrates how to express the volume in terms of \( x \) and \( y \) alone.
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I don't know where to start, can I find out side lengths?View attachment 6538
 

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I would let $h$ be the height of the cuboid...and so the volume $V$ is given by:

$$V=100xyh$$

Now, if $w$ and $\ell$ are the width and length respectively, we know:

$$w=\frac{3y}{h}$$

$$\ell=\frac{\dfrac{x}{3}}{h}$$

Multiplying these two equations together, we obtain:

$$\ell w=\frac{xy}{h^2}$$

Now, the volume is also given by:

$$V=\ell wh$$

Can you now get the volume in terms of $x$ and $y$ alone?
 
V=100(x^2)/2y?
 
Ilikebugs said:
V=100(x^2)/2y?

We have:

$$V=100xyh=\ell wh$$

Thus:

$$100xy=\ell w=\frac{xy}{h^2}\implies h^2=\frac{1}{100}\implies h=\frac{1}{10}$$

Hence:

$$V=100xyh=100xy\cdot\frac{1}{10}=10xy$$
 
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