JEE MAIN - Mathematics (2021 - 27th August Evening Shift - No. 8)
A box open from top is made from a rectangular sheet of dimension a $$\times$$ b by cutting squares each of side x from each of the four corners and folding up the flaps. If the volume of the box is maximum, then x is equal to :
$${{a + b - \sqrt {{a^2} + {b^2} - ab} } \over {12}}$$
$${{a + b - \sqrt {{a^2} + {b^2} + ab} } \over 6}$$
$${{a + b - \sqrt {{a^2} + {b^2} - ab} } \over 6}$$
$${{a + b + \sqrt {{a^2} + {b^2} + ab} } \over 6}$$
Explanation
_27th_August_Evening_Shift_en_8_3.png)
V = l . b . h = (a $$-$$ 2x)(b $$-$$ 2x) x
$$\Rightarrow$$ V(x) = (2x $$-$$ a)(2x $$-$$ b) x
$$\Rightarrow$$ V(x) = 4x3 $$-$$ 2(a + b)x2 + abx
$$ \Rightarrow {d \over {dx}}v(x) = 12{x^2} - 4(a + b)x + ab$$
$${d \over {dx}}(v(x)) = 0 \Rightarrow 12{x^2} - 4(a + b)x + ab = 0 < _\beta ^\alpha $$
$$ \Rightarrow x = {{4(a + b) \pm \sqrt {16{{(a + b)}^2} - 48ab} } \over {2(12)}}$$
$$ = {{(a + b) \pm \sqrt {{a^2} + {b^2} - ab} } \over 6}$$
Let $$x = \alpha = {{(a + b) + \sqrt {{a^2} + {b^2} - ab} } \over 6}$$
$$\beta = {{(a + b) - \sqrt {{a^2} + {b^2} - ab} } \over 6}$$
Now, $$12(x - \alpha )(x - \beta ) = 0$$
_27th_August_Evening_Shift_en_8_2.png)
$$\therefore$$ x = $$\beta$$ $$ = {{a + b - \sqrt {{a^2} + {b^2} - ab} } \over b}$$
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