Question #6d4ae

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Question #6d4ae

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Answer 1 · 2016-10-23T00:12:03

$\frac{2}{3} inch$ $2/3 " inch"$

Explanation:

Let the size of square to be cut from each corner $= x inch$ $=x " inch"$
Volume of the open box so formed $V = l \times w \times h = (8 - 2 x) \times (3 - 2 x) \times x$ $V=lxxwxxh=(8-2 x)xx(3-2x)xx x$
$\Rightarrow V = (24 - 22 x + 4 x^{2}) \times x$ $=>V=(24-22 x+4x^2)xx x$
$\Rightarrow V = 24 x - 22 x^{2} + 4 x^{3}$ $=>V=24x-22 x^2+4x^3$
To find maximum value we need to differentiate $V$ $V$ with respect to $x$ $x$ and equate it to $0$ $0$
$\frac{d V}{d x} = \frac{d}{d x} (24 x - 22 x^{2} + 4 x^{3})$ $(dV)/dx=d/dx(24x-22 x^2+4x^3)$ , setting it equal to zero we get
$24 - 44 x + 12 x^{2} = 0$ $24-44 x+12x^2=0$ , dividing both sides by $4$ $4$ and rearranging we get
$3 x^{2} - 11 x + 6 = 0$ $3x^2-11x+6=0$
Using split the middle term method to find the roots of above we get
$3 x^{2} - 9 x - 2 x + 6 = 0$ $3x^2-9x-2x+6=0$
$\Rightarrow 3 x (x - 3) - 2 (x - 3) = 0$ $=>3x(x-3)-2(x-3)=0$
$\Rightarrow (x - 3) (3 x - 2) = 0$ $=>(x-3)(3x-2)=0$
We get the roots as
$x = 3, \frac{2}{3}$ $x=3, 2/3$

For $x = 3$ $x=3$ , we see that width $w$ $w$ of the box becomes negative. Hence, ignoring this root we have size of the square $\frac{2}{3} inch$ $2/3 " inch"$

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Question #6d4ae

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