I suspect you are missing something in the problem statement but
the base seems to be square, 2 by 2 so the base area = 4
then the volume v is 1/3 *the area of the base * the height
v = (1/3)4 h
I assume the surface area you mean is the area of the sides, not including the base.
Look at one of the sloping sides. Its base is 2 but we need to find its altitude by using trig. It is the hypotenuse of a triangle whose base is 1 and height is h
therefore the altitude is sqrt(1+h^2)
therefore the area of one side is
(1/2)(2)(sqrt(1+h^2))
and the total area of all four sides is
4 sqrt(1+h^2)
so v = 4/3 h
the minimum is of course when h = 0
a = 4 (1+h^2)^.5
da/dh = 4(.5)(1+h^2)^-.5 * 2h
this is clearly zero when h = 0
looking at a square pyramid, if the size of the base remains the same and the height of the pyramid varies what are the minimum surface area and minimum volume the pyramid can have? The base is showing 2 in and the line on the right side of the base which I believe is also still the bottom of the pyramid shows 2 in. the height indicator inside the pyramid has the right angle marking on it but there is no number listed for that height.
I am confused on how to do surface area and volume if V=1/3bh and we don't know the height? Same with trying to figure out the surface area?
1 answer
1 answer