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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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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 in cms to be cut off so that the volume of the box is maximum ?

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1 Answer

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Toolbox:
  • Volume of the box=$l\times b\times h$
  • $\large\frac{d}{dx}$$(xy)=x.\large\frac{d}{dx}$$(y)+y.\large\frac{d}{dx}$$(x)$
Step 1:
Side of the square $\rightarrow (45-2x)(24-2x)x$
Volume of the box=$l\times b\times h$
$V=(45-2x)(24-2x).x$
$\;\;=2x(12-x)(45-2x)$
$\;\;=2x(540-45x-24x+2x^2)$
$\;\;=2x[2x^2-69x+540]$
$\;\;=2(2x^3-69x^2+540x)$
Differentiating with respect to x we get
$\large\frac{dV}{dx}$$=2(6x^2-138x+540)$
$\quad=2\times 6(x^2-23x+90)$
$\quad=12(x^2-23x+90)$
Step 2:
For maxima & minima
$\large\frac{dV}{dx}$$=0$
$12(x^2-23x+90)=0$
$x^2-23x+90=0$
$x^2-5x-18x+90=0$
$x(x-5)-18(x-5)=0$
$(x-18)(x-5)=0$
$x=18,5$
Step 3:
But $x$ cannot be greater than 12
$\Rightarrow x=5$
$\large\frac{d^2V}{dx^2}$$=12(2x-23)$------(1)
Substitute the value of $x=5$ in (1)
$\large\frac{d^2V}{dx^2}=$$12(2\times 5-23)$
$\qquad=12(10-23)$
$\qquad=12(-13)$
$\Rightarrow$ -ve
$\therefore V$ is maximum at $x=5$
(i.e) Square of side 5 cm is cut off from each corner.
answered Aug 8, 2013 by sreemathi.v
edited Aug 19, 2013 by sharmaaparna1
 

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