Respuesta :
Answer:
Therefore the rate change of the diagonal of the foam box is [tex]\frac{2}{\sqrt{21}}[/tex] 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= [tex]\sqrt{l^2+w^2+h^2}[/tex]
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
[tex]\frac{dl}{dt}= 4[/tex] cm per min., [tex]\frac{dw}{dt}=3[/tex] cm per min and [tex]\frac{dh}{dt}=-2[/tex] cm per min[ since the height is decreasing]
∴d= [tex]\sqrt{l^2+w^2+h^2}[/tex]
Differentiating with respect to t
[tex]\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})[/tex]
[tex]\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})[/tex]
Putting [tex]\frac{dl}{dt}= 4[/tex],[tex]\frac{dw}{dt}=3[/tex] , [tex]\frac{dh}{dt}=-2[/tex]
[tex]\Rightarrow \frac{dh}{dt}=\frac{1}{\sqrt{l^2+w^2+h^2}}\{l\times 4 +w\times 3+h\times (-2)\}[/tex]
[tex]\Rightarrow \frac{dh}{dt}=\frac{1}{\sqrt{l^2+w^2+h^2}}(4l+3w-2h)[/tex]
Now
[tex]\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)[/tex]
[tex]=\frac{2}{\sqrt{21}}[/tex]
Therefore the rate change of the diagonal of the foam box is [tex]\frac{2}{\sqrt{21}}[/tex] cm per minute.
