Let
x--------> the length side of the square base of the box
y-------> the height of the box
we know that
The surface area of the box is equal to
S A = 4 x y + x 2
S A = 10 , 800 c m 2
so
10 , 800 = 4 x y + x 2
y = ( 10 , 800 − x 2 ) / ( 4 x )
y = ( 2 , 700 − 0.25 x 2 ) / ( x ) --------> equation A
the volume of the box is equal to
V = x 2 y --------> equation B
Substitute the equation A in the equation B
V = x 2 ∗ ( 2 , 700 − 0.25 x 2 ) / ( x )
V = x ∗ ( 2 , 700 − 0.25 x 2 )
V = ( 2 , 700 x − 0.25 x 3 )
using a graphing tool
see the attached figure
For x = 60 c m
V o l u m e = 108 , 000 c m 3
the point ( 60 , 108 , 000 ) is a maximum of the function
Find the dimensions of the box
x = 60 c m
Find the value of y
V = x 2 y
y = V / x 2
y = 108 , 000/6 0 2
y = 30 c m
The dimensions of the box are
60 c m ∗ 60 c m ∗ 30 c m
The largest possible volume of the box is
108 , 000 c m 3
The largest possible volume of the box is 108000 c m 3 . ;
The largest possible volume of a box with a square base and an open top, given a surface area of 10,800 cm², is 108,000 cm³. The dimensions of the box are 60 cm by 60 cm for the base and 30 cm for the height. This was determined by optimizing the volume based on the surface area constraint.
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