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Sagot :
The maximum volume of the box in terms of A will be √3A/4 -3√3/4 cu cm.
It is given that the box has an open top and a square base.
This implies length = breadth.
Let them be x cm
Now, the area of the box is A sq cm
This means
lb + 2bh + 2lh = A
or, x² + 2xh + 2xh = A
or, 4xh + x² = A
or, h = A/4x - x/4
Now volume (V) = lbh
= x²(A/4x - x/4)
= Ax/4 - x³/4
Now to maximize volume we will first differentiate with respect to x and equate it to zero. Hence we get
V' =0
or, A/4 - 3x²/4 = 0
or 3x²/4 = A/4
or, x² = 3
or, x = √3
Now t check whether this gives us the maximum volume we will calculate V"(√3)
V" = -3x/2
V"(√3) = -3√3/2
Since V"(√3) is negative we will get maximum value for x = √3
Now the volume is
A√3/4 - (√3)³/4
= √3A/4 -3√3/4 cu cm
To know more about maximization visit
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