18) A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum ?

It is given that the sides of the rectangle are 45 cm and 24 cm
Let assume  the side of the square to be cut off is x cm
Then,
Volume of cube $V(x) = x(45-2x)(24-2x)$
$V^{'}(x) = (45-2x)(24-2x) + (-2)(x)(24-2x)+(-2)(x)(45-2x)\\$
$1080 + 4x^2 - 138x - 48x + 4x^2 - 90x +4x^2\\ 12x^2 - 276x + 1080$
$V^{'}(x) = 0\\ 12(x^2 - 23x+90)=0\\ x^2-23x+90 = 0\\ x^2-18x-5x+23=0\\ (x-18)(x-5)=0\\ x =18 \ and \ x = 5$
But  x cannot be equal to 18 because then side (24 - 2x) become negative which is not possible so we neglect value x= 18
Hence, x = 5 is the critical value
Now,
$V^{''}(x)=24x-276\\ V^{''}(5)=24\times5 - 276\\ V^{''}(5)= -156 < 0$
Hence, x  = 5 is the point of maxima
Hence,  the side of the square to be cut off is 5 cm  so that the volume of the box is maximum

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