Asked by Dodie
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.
Answers
Answered by
Reiny
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
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