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zeesbrat3
 one year ago
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. Find the rate at which the volume of the box is changing when the edge length of the base is 10 cm, the edge length of the base is increasing at a rate of 3 cm/min, the height of the box is 5 cm, and the height is decreasing at a rate of 1 cm/min.
zeesbrat3
 one year ago
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. Find the rate at which the volume of the box is changing when the edge length of the base is 10 cm, the edge length of the base is increasing at a rate of 3 cm/min, the height of the box is 5 cm, and the height is decreasing at a rate of 1 cm/min.

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zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.0Please help and explain

butterflydreamer
 one year ago
Best ResponseYou've already chosen the best response.0write out all the information given to us first \[V=x^2h\] and we want to find dV/dt = ? when \[x = 10, \frac{ dx }{ dt} = 3 \] \[h = 5, \frac{ dh }{ dt }= 1\] Okaay and now you want to differentiate V= x^2h in terms of both "x" and "h" like this: \[V = x^2h \rightarrow \frac{ dV }{ dx }= 2xh \] \[V = x^2h \rightarrow \frac{ dV }{ dh } = x^2 \] Finally, use this formula to work out the rate of change of the volume :) \[\frac{ dV }{ dt } = \frac{ dV }{ dx } \times \frac{ dx }{ dt} + \frac{ dV }{ dh } \times \frac{ dh }{dt }\] Now scroll up and just plug in all the information from the above and we get: \[\frac{ dV }{ dt} = 2xh \times 3 + x^2 \times 1 \rightarrow \frac{ dV }{ dt } = 2(10)(5) \times 3 + (10)^2 \times 1 \] and you should be able to solve from here :)
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