Hi
You will find a similar problem here
http://www.mathhelpforum.com/math-he...ted-rates.html
Please check the attachment. Now I only know how to find the volume of the pyramid, that's 1/3 Ah and surface area is sum of area of triangles + sum of base square.
I think I can do the differentiation part in the question but I need help with the earlier part.
Hi
You will find a similar problem here
http://www.mathhelpforum.com/math-he...ted-rates.html
Note to other users: The text of the exercise, as displayed in the graphic, is quoted below.
The image shows the corners of the smaller bottom base of the frustum of the pyramid as being ABCD, with A on the left-hand front corner and B on the right-hand front corner. The larger upper base is EFGH, with the left-hand front corner being E and the right-hand front corner being F.An underground oil storage tank ABCDEFGH is part of an inverted square pyramid, as shown.
What have you tried? How far did you get? Where are you stuck?The complete pyramid has a square base of side 12 meters and height 18 meters. The depth of the tank is 12 meters.
When the depth of the oil in the tank is meters, show that the volume V, in cubic meters, is given by:
. . . . .
Oil is being added to the tank at a constant rate of 4.5 cubic meters per second. At the moment when the depth of oil is 8 meters, find the rate at which the depth is increasing.
You found the formula for the area of a pyramid with base area 144 and height 18. You used similar triangles to find the lengths of the sides of the lower base (and thus of the cut-off pyramid). You used this to find the volume of the cut-off pyramid. You used similar triangles again to find the volume of the pyramid with a height of 6 + h (the "6" being the cut-off portion and the "h" being the height of the oil in the remaining portion). And... then what?
Please be complete. Thank you!