Calculus - Luggage with maximum volume 

 Some airlines have restrictions on the size of items of luggage that passengers are allowed to take with them. Suppose that one has a rule that the sum of the length, width and height of any piece of luggage must be less than or equal to 150 cm. A passenger wants to take a box of the maximum allowable volume. If the length and width are to be equal, what should the dimensions be?
length = width = 

height = 

In this case, what is the volume?
volume = 

(for each, include units)

If the length is be twice the width, what should the dimensions be?
length = 

width=

height = 

In this case, what is the volume?
volume = 

1 Answer

Let $x,y,z$ be the length, width, and height of the luggage. We would like to maximize the colume
\[v=xyz\]
subject to the constraint that 
\[x+y+z=150.\]


(i) If the length and width are to be equal, then $y=x$. So we have
\[v=x^2z   \text{and}   2x+z=150  \Rightarrow  z=150-2x.\]
Hence 
\[v=x^2(150-2x)=150x^2 -2x^3.\]
\[v'=300x-6x^2=0    \Rightarrow 6x(50-x)=0.\]
Thus we must have $x=50$ to maximize the volume. Therefore $x=y=50$: 

length= width = 50 cm. 

height = z=150-2x=150 -2*50=50 cm.

In this case, what is the volume?
volume = xyz= (50)^3=125000 cm$^3$.


If the length is be twice the width, then $x=2y$. So we have
\[v=(2y)yz=2y^2z \]
\[x+y+z=2y+y+z=150  \Rightarrow  z=150-3y.\]
Hence 
\[v=2y^2(150-3y)=300y^2 -6y^3.\]
\[v'=600y-18y^2=0    \Rightarrow 6y(100-3y)=0.\]
Therefore $y=\frac{100}{3}$. 



length = $x=2y= \frac{200}{3}$ cm

width= $y=\frac{100}{3}$

height = $z=150-3y=150- \frac{300}{3}=50$ cm

In this case, what is the volume?
volume $= xyz=\frac{200}{3}* \frac{100}{3}* 50=\frac{100000}{9}$ cm

Erdos Erdos
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