Question

a box with square base of length x and height y is changing shape such that \\( \frac { d x } { d t } = 3 \mathrm { cm } / \mathrm { s }, \frac { d y } { d t } = 7 \mathrm { cm } / \mathrm { s } \\). find the rate of change of volume when \\( x = 2 \mathrm { cm } \\), \\( y = 4 \mathrm { cm } \\).
recall: the volume v of a box with length \\( \ell \\), width w, and height h is \\( v = \ell w h \\).
\\( 44 \mathrm { cm } ^ { 3 } / \mathrm { s } \\)
\\( 89 \mathrm { cm } ^ { 3 } / \mathrm { s } \\)
\\( 76 \mathrm { cm } ^ { 3 } / \mathrm { s } \\)
\\( 31 \mathrm { cm } ^ { 3 } / \mathrm { s } \\)
\\( 5 \mathrm { cm } ^ { 3 } / \mathrm { s } \\)

Explanation:

Step1: Write the volume formula

The volume \(V\) of the box with square base is \(V = x^{2}y\).

Step2: Differentiate \(V\) with respect to \(t\) using the product rule

The product rule \((uv)^\prime=u^\prime v + uv^\prime\), where \(u = x^{2}\) and \(v = y\).
First, find \(\frac{dV}{dt}\):
\(\frac{dV}{dt}=\frac{d(x^{2}y)}{dt}=2xy\frac{dx}{dt}+x^{2}\frac{dy}{dt}\)

Step3: Substitute the given values

Given \(x = 2\mathrm{cm}\), \(y = 4\mathrm{cm}\), \(\frac{dx}{dt}=3\mathrm{cm/s}\), \(\frac{dy}{dt}=7\mathrm{cm/s}\)
\(\frac{dV}{dt}=2\times2\times4\times3+2^{2}\times7\)
\(=48 + 28\)
\(=76\mathrm{cm^{3}/s}\)

Answer:

\(76\mathrm{cm^{3}/s}\) (the third option)