Answer:
The size square removed from each corner = 32.15 cm²
Step-by-step explanation:
The volume of the box = Length * Breadth * Height
Let r be the size removed from each corner
Note that at maximum volume, [tex]\frac{dV}{dr} = 0[/tex]
The original length of the cardboard is 119 cm, if you remove a size of r (This typically will be the height of the box) from the corner, since there are two corners corresponding to the length of the box, the length of the box will be:
Length, L = 119 - 2r
Similarly for the breadth, B = 24 - 2r
And the height as stated earlier, H = r
Volume, V = L*B*H
V = (119-2r)(24-2r)r
V = r(2856 - 238r - 48r + 4r²)
V = 4r³ - 286r² + 2856r
At maximum volume dV/dr = 0
dV/dr = 12r² - 572r + 2856
12r² - 572r + 2856 = 0
By solving the quadratic equation above for the value of r:
r = 5.67 or 42
r cannot be 42 because the size removed from the corner of the cardboard cannot be more than the width of the cardboard.
Note that the area of a square is r²
Therefore, the size square removed from each corner = 5.67² = 32.15 cm²
