Mathematics: The volume of a cube is increasing at a constant rate of 459 cubic inches per second. At the instant when the volume of the cube is 408 cubic inches, what is the rate of change of the surface area of the cube? Round your answer to three decimal places (if necessary).

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The volume of a cube is increasing at a constant

natishtaylor1p8dirz

16.03.2022

$\frac{ds}{dt} =31.952$

Step-by-step explanation:

From the question we are told that

Cube increase rate 459 cubic inches

At 408

Generally the expressions of the is mathematically given to be

$\frac{dv}{dt} =459$

$\frac{dv}{dt} =3*l^2 \frac{dl}{dt}$

v=408

Generally

v=l^3

Therefore solving Mathematically

$\frac{dv}{dt} =3*l^2 \frac{dl}{dt}$

$459 =3*(7.416859539)^2 \frac{dl}{dt}$

$459 =165.0294163\frac{dl}{dt}$

$\frac{dl}{dt} =\frac{165.0294163}{459}$

$\frac{dl}{dt} =0.359$

Therefore solving mathematically for surface area

Surface area S=6l^2

$\frac{ds}{dt} =6(2l* \frac{dl}{dt})$

$\frac{ds}{dt} =12l* \frac{dl}{dt}$

$\frac{ds}{dt} =12*(7.416859539)*0.359$

$\frac{ds}{dt} =31.95183089$

$\frac{ds}{dt} =31.952$

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