Question

A rectangular box has a square base with an edge length of x cm and a height of h cm. The volume of the box is given by V = x2h cm3. Find the rate at which the volume of the box is changing when the edge length of the base is 12 cm, the edge length of the base is decreasing at a rate of 2 cm/min, the height of the box is 6 cm, and the height is increasing at a rate of 1 cm/min.

Answer 1

It is given in the question that

A rectangular box has a square base with an edge length of x cm and a height of h cm. The volume of the box is given by

$V = x^2h cm^3$

And the edge length of the base is 12 cm, the edge length of the base is decreasing at a rate of 2 cm/min, the height of the box is 6 cm, and the height is increasing at a rate of 1 cm/min.

Here we differentiate V with respect to t, and we use product rule, that is

$V = 2xh(dx/dt)+ x^2(dh/dt)$

Substituting the given values , we will get

$[tex]dV/dt = 2(12)(6)(-2)+ 12^2(1)$[/tex]

$dV/dt = -288+144 = -144cm^3/min$

So at that moment, the volume is decreasing at the rate of

$144 cm^3/min$

Answer 2

The volume is decreasing at the rate of $\boxed{{\mathbf{144c}}{{\mathbf{m}}^{\mathbf{3}}}{\mathbf{/min}}}$ .

Further explanation:

Given:

It is given that a rectangular box a square base with a length of the side is $x{\text{ cm}}$  and has a height of $h{\text{ cm}}$ .

Step by step explanation:

Step 1:

The volume of the rectangular box can be calculated as,

$V=lbh$

Here, $l$  is the length,  $b$ is breadth,  $h$ is the height.

Therefore, volume of the box can be calculated as,

$V={x^2}h{\text{ c}}{{\text{m}}^{\text{3}}}$

Step 2:

Now differentiate volume with respect to $l$  both sides by the use of product rule.

  $\frac{{dV}}{{dt}}=2xh\left({\frac{{dx}}{{dt}}}\right)+{x^2}\left({\frac{{dh}}{{dt}}}\right)$

It is given that the length of the base is $12{\text{ cm}}$  and it decreasing at the rate of .

The height of the box is $6{\text{ cm}}$  and the height is increasing at the rate of $1{\text{ cm/min}}$ .

Therefore, the values are $x=12,h=6,\frac{{dx}}{{dt}}=-2{\text{ and}}\frac{{dh}}{{dt}}=1$

Now substitute the values  $x=12,h=6,\frac{{dx}}{{dt}}=-2{\text{ and}}\frac{{dh}}{{dt}}=1$ in the differential equation.

$\begin{gathered}\frac{{dV}}{{dt}}=2xh\left({\frac{{dx}}{{dt}}}\right)+{x^2}\left({\frac{{dh}}{{dt}}}\right)\hfill\\\frac{{dV}}{{dt}}=2\left({12}\right)\left(6\right)\left({-2}\right)+{\left({12}\right)^2}\left(1\right)\hfill\\\frac{{dV}}{{dt}}=-144{\text{c}}{{\text{m}}^{\text{3}}}/\min\hfill\\\end{gathered}$

Therefore, it can be seen that the volume is decreasing as the value of $\frac{{dV}}{{dt}}$  involves negative sign.

Thus, the volume is decreasing at the rate of $144c{m^3}/\min$ .

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Answer details:

Grade: Middle school

Subject: Mathematics

Chapter: Application of derivative

Keywords: side, length, square, base, differentiate, decreasing, increasing, rectangular box, respect, product rule, height, rate, volume of the box