let the length, the width and the height be
a , b, and c respectively
make a sketch to see that
4a + 4b + 4c = 76 ---> a+b+c = 19
or c = 19 - a - b
2ab + 2ac + 2bc = 206 ---> ab + ac + bc = 103
abc = 165
or c = 165/(ab)
19-a-b = 165/ab
19ab - a^2 b - ab^2 = 165
ab(19 - a^2 - b^2) = 165
from ab+ac+bc=103
c(a+b) = 103 - ab
c = (103-ab)/(a+b)
165/(ab) = (103-ab)/(a+b)
165a + 165b = 103ab - a^2 b^2
165(a+b) = ab(103 - ab)
ok, going on a hunch that the numbers are whole numbers, let's look at abc = 165
and 165 = 11x5x3 , all primes
test for ab+bc +ac = 55 + 15 + 33 = 103
a+b+c = 11+5+3 = 19
well, that was a lucky hunch,
the sides are 11 , 5, and 3
Find the dimensions of a rectangular solid for which the sum of all the edges is 76, the total surface area is 206 square units, and the volume is 165 cubic units.
1 answer