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shaqadry
Let A be the total surface area of a rectangular container of square base and height, h. At a certain instant, the surface area is decreasing at 24 cm^2/s while the width of the base is 5 cm and increasing at 2 cm/s. Determine whether the height is increasing or decreasing at that instant if the height is 3 cm. (Ans: dh/dt = -4.4 cm/s) Somebody PLEAAAAAASSSEEE explain this to me. :(
This appears to be a related rate problem, but the answer tells you that the height is decreasing (note the - in the -4.4cm/sec.
okay :/ how do i solve it?
Where did the answer come from? If that is your answer, then you have already solved it.
got it from the lecturer. but i have no idea how to solve it.
O.K. First you need to express the total surface area (A) in terms of the base and height. The base is square (given) so assign x as the dimension of the base. The surface area of the base is then x^2 (x squared). This is not the total A, just the base area. The area of the rest of the container is to be expressed in terms of x (base dimension) and h (height). There are 4 sides each having an area hx, so that equals 4hx. Now the total surface area A is:\[x ^{2}+4hx\]. You must understand this before proceeding. Do you?
so the top of the rectangle doesnt count?
Do you understand rate of change or derivatives? No it is a container, no mention of a lid, but if it did count the terms which is now x^2 would become 2x^2. we can work it both ways, because it might be a closed container.
okay i understand. so then we find dA/dh? and the x remains? after that what?
A'=-24 (cm)^2 given You are also given x' as 2cm/s you are given x (x = 5) you are given h (h=5) solve for h'
I am kind of rusty here on implicit differentiation, hopefully you are current.
I come up with -3.666 not the approved answer. I will try with the lid!
thank you so much! appreciate your help :)
With the lid I get -5.55 somewhere I've erred, suggest going over it or did you get the -4.4