Arectangular box with a volume of 960 ftcubed3 is to be constructed with a square base and top. the cost per square foot for the bottom is 15cents¢, for the top is 10cents¢, and for the sides is 1.5cents¢. what dimensions will minimize the cost?

Yas

Square base and top, means L=W

so

V=HL²
V=960

bottom cost=15L²

top cost=10L²

sides=2H(L+W)1.5 or 3H(2L) or 6HL

so
total cost=15L²+10L²+6HL or
TC=25L²+6HL

eliminate height
we know
960=HL²
divide both sides by L²
960/L²=H
sub that for H
 

TC=25L²+6(960/L²)L
TC=25L²+5760/L
find where TC is minimum

take the derivitive
50L-5760/L²
or
(10)(5L³-576)/(L²)
set numerator to zero
5L³-576=0
solve
L=0.8∛225

H=960/L²
H=(20/3)∛225

if we were to sub,we would find that minimum cost is 720(∛15) or aprox 1775.67 cents or $17.76

dimentions

L=0.8∛225 aprox 4.86576 ft
W=0.8∛225 aprox 4.86576 ft
H=(20/3)∛225 aprox 40.548 ft
min cost is $17.76

No a triangle cannot be formed because when you add the two smallest number together (which would be 11) the have to be bigger than the number it is "against" ( which would be 12)