Some airlines have restrictions on the size of items of luggage that passengers

Question

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 210 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 =  
(for each, include units)

Explanation / Answer

Let length of the luggage=x=widthof the luggage

Height of the luggage=y

SInce the luggage is a rectangular box and given, sum of the lengths must be equal to or less than 210

so, x + x + y = 210

==> 2x + y = 210 ==> y=210-2x

And volume of rectangular box=V=x*x*y=x^2y=x^2(210-2x)=210x^2-2x^3

Differentiation the volume

dV/dx = 420x - 6x^2.

For Maximum volume, dV/dx=0

==>420x-6x^2=0   ==>x(420-6x)=0, x=0 and x=420/6=70 unit,

so, y=210-2*70=210-140=70 unit
Hence, x=y=70 unit

that is lenght=x=70 unit

width=x=70 unit

height=y=70 unit

Volume=V=x^2y=(70)^2*70=343,000 unit^3

Case 2:- If the length is twice the width, let width=x, length=2x, height=y

and x+2x+y=210 ==> y=210-3x

so, Volume v=x*2x*y=2x^2*y=2x^2*(210-3x)=420x^2-6x^3=V

=> dV/dx=840x-18x^2

For Maximum, dV/dx=0 => x(840-18x)=0   => x=840/18=140/3=width

and length=2x=280/3 unit

Height=210-(280/3+140/3)=70 unit

So, volume=V=2x^2*y=2*(140/3)^2*70=2744000/9 unit^3

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.