Calculus problem

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 216 cm. A passenger wants to take a box of the maximum allowable volume.

(a) If the length and width are to be equal, what should the dimensions be? (For each answer, include units)

length = width = 

height = 

In this case, the maximum volume is 

(b) If the length is be twice the width, what should the dimensions be? (For each answer, include units)

length = 

width = 

height = 

In this case, the maximum volume is 

Answer 1

Hi again,

(a)So let's follow what the question is saying.

the sum of the length, width and height of any piece of luggage must be less than or equal to 216 cm

L+W+H=216 (since we want the max volume, we can ignore the "less than part" and set them equal)

If the length and width are to be equal

L=W

Let's substitute L with W, we will get this new equation

2W+H = 216

arrange H to the left

H=216-2W

So what is volume

V=L*W*H=W^2*(216-2W)=216W^2-2W^3

Similar to the last question you asked, when you want to maximize something, you find the first derivative and set it equal to 0 .

dV/dW=432W-6W^2

432W-6W^2=0

divide 6 on both side

72w-w^2=0

W=72

Plug it back in, you will find W=L=H=72 cm.

Therefore maximum volume = 72^3=373248 cm^3 UNITS!!

Ps: Interestingly, most questions like this, the three dimensions always equal to each other. Just in case when you can't finish your work during an exam, this will be a life saver lol.

b) the only thing changed is L=2W now.

Given L+W+H=216

Now we know 3W+H=216

V=L*W*H= 2W*W*(216-3W)=432W^2-6W^3

V prime = 864W-18W^2

Set V prime = 0 and solve it by dividing both sides by 18

we will get W = 48

Plug W back in there, you will get W=48 cm, L=96 cm, and H= 72 cm

Therefore maximum volume = 48cm*96cm*72cm=331776 cm^3 UNITS....

hope it helps!