Asked by Mohammed
Find the most economical proportions for a covered box of fixed volume whose base is a rectangle with one side three times the other.
Answers
Answered by
helli
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
$V = x \, (3x) \, y
V = 3x^2 \, y$
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
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