Find the most economical proportions for a covered box of fixed volume whose base is a rectangle with one side three times the other.

Answer 1

Given Volume:
$V = x \, (3x) \, y
V = 3x^2 \, y$

0=3x2y′+6xy
0=3x2y′+6xy

y′=−2y/x
y′=−2y/x



Total Area:
AT=2(3x2)+2(3xy)+2(xy)
AT=2(3x2)+2(3xy)+2(xy)

AT=6x2+8xy
AT=6x2+8xy

dAT/dx=12x+8(xy′+y)=0
dAT/dx=12x+8(xy′+y)=0

12x+8[x(−2y/x)+y]=0
12x+8[x(−2y/x)+y]=0

12x+8[−2y+y]=0
12x+8[−2y+y]=0

12x=8y
12x=8y

y=32x
y=32x



Altitude = 3/2 × shorter side of base. answer