Q.147. Maxima/Minima

Question 147.
A rectangular sheet of metal measures 8cm by 3cm. From each corner, a square of x cm is removed and the flaps so formed are bent up to make a small open box. Find the maximum volume of the box.

Answer 147.
When x cm * x cm squares are removed from four corners,
the rectangular base remaining will be (8 - 2x) * (3 - 2x) cm^2
The height of the box will be = x cm
=> Volume,
V = x * (8 - 2x) * (3 - 2x)
= 4x^3 - 22x^2 + 24x cm^3.

For V to be maximum, dV/dx = 0 and d^2V/dx^2 < 0
dV/dx = 0
=> 12x^2 - 44x + 24 = 0
=> 3x^2 - 11x + 6 = 0
=> (3x - 2)(x - 3) = 0
=> x = 2/3 cm or 3 cm
x = 3 cm is not possible as 3x3 squares cannot be cut from a sheet of 8x3
=> x = 2/3 cm

Now, d^V/dx^2 = 24x - 44
For x = 2/3, d^2V/dx^2 = 24*(2/3) - 44 < 0
=> for x = 2/3 cm, the volume of the box will be maximum
and the maximum volume
= 4(2/3)^3 - 22(2/3)^2 + 24(2/3) cm^3
= (32/27 - 88/9 + 48/3) cm^3
= (32 - 264 + 432) / 27 cm^3
= 200/27 cm^3
≈ 7.41 cm^3.

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