CBSE BoardEnglish MediumSTD 12 ScienceApplied MathsDifferentiation and it's Applications5 Marks
Question
Show that the surface area of closed cuboid with square base and given volume is minimum when it is a cube.
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Answer
Let the length and breadth be $x$ units (square base), height of cuboid be $h$ unit. Volume of cuboid $(v)=x^2 h$ $\therefore h=\frac{v}{x^2}$. Surface area of cuboid$(s)=2\left(x^2+x h+x h\right)$ $\therefore s=2 x^2+4 x h$ $\Rightarrow s=2 x^2+4 x \times \frac{v}{x^2}$ $\Rightarrow s=2 x^2+4 v \frac{1}{x}$ $\therefore \frac{d s}{d x}=4 x-4 v \frac{1}{x^2}$ For maximum and minimum surface, $\frac{d s}{d x}=0$ $\therefore 4 x-4 v \cdot \frac{1}{x^2}=0$ $\Rightarrow 4 x=\frac{4 v}{x^2}$ $\Rightarrow \quad x^3=v$ $\Rightarrow x^3=x^2 h$ $\Rightarrow \quad x=h$ Now, $\frac{d^2 s}{d x^2}=4+\frac{8 v}{x^3}$ At $x=h, \frac{d^2 s}{d x^2}=4+\frac{8 v}{h^3}>0$ $\therefore$ Surface area is minimum, when $x=h$. i.e., when $x=h$, it is a cube, surface area will be minimum. x = h Hence proved.
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