+2668 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!!
+66 \(𝑥𝑦𝑧=720 \\ 𝑥𝑦+𝑥𝑧+𝑦𝑧=333 \),
Im not sure what to do from here, we could make \(𝑥≥𝑦≥𝑧\), but I don't see how that helps
+2668 I did that, and I just brute forced it from here...
Surely there's a better way...
+2668 No, but I found the answer already by guessing and checking, \(x = 16\), \(y = 15\) and \(z = 3\)
+66 I found online that you can do
\(720≥z^3 \\ 1≤𝑧≤8 \\ \),
Then you try out 8 cases
+118687 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
