Question

solve questions number 6 plz

Answer 1

The volume: V=x³. The surface: S=6x².

Given the surface is 216cm², we can find the side:

$\\ \sf ~~~\implies 6x² = 216$

$\sf ~~~\implies x² = 36$

$\sf ~~~\implies x = \sqrt{36}$

$\sf ~~~\implies x = 6$

• Given the volume increases at the rate 24cm³, we can find at what rate the side is increases:

$\dfrac{dV}{dt} \sf=3x^2 × \dfrac{dx}{dt} = 24 \implies\dfrac{dx}{dt} = \dfrac{24}{3×{6}^2 } = \dfrac{2}{9}$

• Given the volume and surface equation, we can differentiate the volume:

$\sf V = s × \dfrac{x}{ 6}$

${\implies\sf \dfrac{dV}{dt} = \dfrac{ds}{dt} \times \dfrac{x}{ 6} + s \times \dfrac{1}{6} \times\dfrac{dx}{dt} = 24}$

${ \implies\dfrac{dS}{dt} \sf× \dfrac{6}{6} + 216 × \dfrac{1}{6} × \dfrac{2}{9} = 24}$

${ \implies \bf\dfrac{dS}{dt} =16}$

• At 16cm²/min fast surface area increasing when the surface area is 216 cm².

Answer 2

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The volume: V=x³. The surface: S=6x².

Given the surface is 216cm², we can find the side:

$\begin{gathered} \end{gathered} </p><p> ⟹6x²=216 \\ </p><p> </p><p> </p><p></p><p>\sf \implies x² = 36 \\ </p><p> </p><p></p><p> ⟹x=6</p><p> </p><p> </p><p>$

Given the volume increases at the rate 24cm³, we can find at what rate the side is increases:

$\dfrac{dV}{dt} \sf=3x^2 × \dfrac{dx}{dt} = 24 \\ \\ \implies\dfrac{dx}{dt} = \dfrac{24}{3×{6}^2 } = \dfrac{2}{9} </p><p>dt</p><p>dV</p><p> </p><p> =3x </p><p>2</p><p> × </p><p>dt</p><p>dx</p><p> </p><p> =24 \\ ⟹ </p><p> dt</p><p>dx</p><p> </p><p> = </p><p>3×6 </p><p>2</p><p> </p><p>24</p><p> </p><p> = </p><p>9</p><p>2</p><p> </p><p> </p><p> \\ Given \: \: the \: \: volume \: \: and \: \: surface \: \: equation, \: \: we \: \: can \: \: differentiate \: \: the \: \: volume: \: \: </p><p>\sf V = s × \dfrac{x}{ 6}V=s× </p><p>6</p><p>x</p><p> </p><p> </p><p></p><p> \\ {\implies\sf \dfrac{dV}{dt} = \dfrac{ds}{dt} \times \dfrac{x}{ 6} + s \times \dfrac{1}{6} \times\dfrac{dx}{dt} = 24} \\ \\ ⟹ </p><p>dt</p><p>dV</p><p> </p><p> = </p><p>dt</p><p>ds</p><p> </p><p> × </p><p>6</p><p>x</p><p> </p><p> +s× </p><p>6</p><p>1</p><p> </p><p> × </p><p>dt</p><p>dx</p><p> </p><p> =24</p><p></p><p> \\ \\ { \implies\dfrac{dS}{dt} \sf× \dfrac{6}{6} + 216 × \dfrac{1}{6} × \dfrac{2}{9} = 24} \\ \\ ⟹ </p><p>dt</p><p>dS</p><p> </p><p> × </p><p>6</p><p>6</p><p> </p><p> +216× </p><p>6</p><p>1</p><p> </p><p> × </p><p>9</p><p>2</p><p> </p><p> =24</p><p></p><p>\begin{gathered} { \implies \bf\dfrac{dS}{dt} =16} \\ \\ \end{gathered} </p><p>⟹ </p><p>dt</p><p>dS</p><p> </p><p> =16</p><p> </p><p> </p><p>$

At 16cm²/min fast surface area increasing when the surface area is 216 cm².

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