A rectangular prism has a volume of \(720cm^3\) and a surface area of \(666cm^3\). Its edges are integers. What are the side lengths of the prism?

I solved this problem the "dumb" way, by trying out every case, but I wonder if there is a "better", more algebraic way to solve it.

Thanks for any help!!

BuilderBoi Apr 17, 2022

#1**0 **

\(𝑥𝑦𝑧=720 \\ 𝑥𝑦+𝑥𝑧+𝑦𝑧=333 \),

Im not sure what to do from here, we could make \(𝑥≥𝑦≥𝑧\), but I don't see how that helps

qjin27 Apr 17, 2022

#2**+1 **

I did that, and I just brute forced it from here...

Surely there's a better way...

BuilderBoi
Apr 17, 2022

#6**+1 **

No, but I found the answer already by guessing and checking, \(x = 16\), \(y = 15\) and \(z = 3\)

BuilderBoi
Apr 17, 2022

#3**+1 **

I found online that you can do

\(720≥z^3 \\ 1≤𝑧≤8 \\ \),

Then you try out 8 cases

qjin27 Apr 17, 2022

#8**+3 **

Here is how I approached it

\(xyz=720\\ xy+xz+yz=333\\ \sqrt[3]{333}\approx 8.9\)

this means that at least one of the dimensions is 8 or less, and at least one is 9 or more

Factor 720 and you get

\(720=2^4*3^2*5\)

the factors 8 and smaller are 1,2,3,4,5,6,8

Try x=8

If x=8 then yz=720/8 = 90

8y+8z+90=333 so 8y+8z = 333-90 = 243

But 243 is not divisable by 8 so 8 is not one ofthe dimensions.

Try x=6 I ran into the same problem

Try x=5 same problem

Try x=4 same problem

Try x=3

If x=3 then yz= 720/3 = 240

3y+3z+240=333

3y+3z=93 so y+z=31

31=15+16, that sounds promising

If the sides, x,y and z are 3, 15, and 16 then

(3*15)+(3*16)+(15*16) = 45+48+240 = 333 Bingo

the sides are 3,15 and 16

Melody Apr 18, 2022