Question

A foam box's length and width are increasing by 4 and 3 centimeters per minute, respectively, while its height is decreasing by 2 centimeters per minute. At the time when the box's length, width, and height are 1, 2, and 4, respectively, what is the rate of change of the length of the long diagonal of the box

Answer 1

Answer:

Therefore the rate change of the diagonal of the foam box is $\frac{2}{\sqrt{21}}$  cm per minute.

Step-by-step explanation:

Let the length, width and height of the foam be l, w and h respectively.

The diagonal of the rectangle is  d= $\sqrt{l^2+w^2+h^2}$

Given that, A foam box's length and width are increasing by 4 and 3 cm per minute.

Its height is decreasing by 2 cm per minute.

That is

$\frac{dl}{dt}= 4$ cm per min., $\frac{dw}{dt}=3$ cm per min and  $\frac{dh}{dt}=-2$  cm per min[ since the height is decreasing]

∴d= $\sqrt{l^2+w^2+h^2}$

Differentiating with respect to t

$\frac{dh}{dt}=\frac{1}{2\sqrt{l^2+w^2+h^2}}(2l\frac{dl}{dt}+2w\frac{dw}{dt}+2h\frac{dh}{dt})$

$\Rightarrow \frac{dh}{dt}=\frac{1}{\sqrt{l^2+w^2+h^2}}(l\frac{dl}{dt}+w\frac{dw}{dt}+h\frac{dh}{dt})$

Putting $\frac{dl}{dt}= 4$,$\frac{dw}{dt}=3$  ,  $\frac{dh}{dt}=-2$

$\Rightarrow \frac{dh}{dt}=\frac{1}{\sqrt{l^2+w^2+h^2}}\{l\times 4 +w\times 3+h\times (-2)\}$

$\Rightarrow \frac{dh}{dt}=\frac{1}{\sqrt{l^2+w^2+h^2}}(4l+3w-2h)$

Now

$\left\frac{dh}{dt}\right|_{l=1,w=2,h=4}=\frac{1}{\sqrt{1^2+2^2+4^2}}(4\times 1+3\times2-2\times4)$

                      $=\frac{2}{\sqrt{21}}$

Therefore the rate change of the diagonal of the foam box is $\frac{2}{\sqrt{21}}$  cm per minute.