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(C) The volume of the liquid in the vessel is given by $V(x) = x^2(15 - x) = 15x^2 - x^3$,where $x$ is the depth of the liquid.Since the vessel is clo

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$10$

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(C) The volume of the liquid in the vessel is given by $V(x) = x^2(15 - x) = 15x^2 - x^3$,where $x$ is the depth of the liquid.Since the vessel is closed and tapers to a point at both ends $E$ and $F$,the maximum volume of the liquid corresponds to the total capacity of the vessel,which occurs when the vessel is completely full.To find the maximum volume,we differentiate $V(x)$ with respect to $x$:$\frac{dV}{dx} = \frac{d}{dx}(15x^2 - x^3) = 30x - 3x^2$.Setting $\frac{dV}{dx} = 0$ for critical points:$3x(10 - x) = 0 \implies x = 0$ or $x = 10$.We check the second derivative to confirm the maximum:$\frac{d^2V}{dx^2} = 30 - 6x$.At $x = 10$,$\frac{d^2V}{dx^2} = 30 - 6(10) = -30 Since the second derivative is negative at $x = 10$,the volume is maximum when the depth $x = 10 \, 	ext{cm}$.Thus,the total length $EF$ of the vessel is $10 \, 	ext{cm}$.

$A$ closed vessel tapers to a point both at its top $E$ and its bottom $F$ and is fixed with $EF$ vertical. When the depth of the liquid in it is $x \, \text{cm}$,the volume of the liquid in it is $V(x) = x^2 (15 - x) \, \text{cu. cm}$. The length $EF$ is ........ $\text{cm}$.

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