Wll, there is much unsaid here, but if we assume the pyramid has a square base of side length x, and height y, then
V = 1/3 x^2 y = 2.2
y = 6.6/x^2
Now, the area of tent is
x^2 for the floor
2x?(x^2+4y^2) for the sides
So, you want to minimize the cost
c = x^2 + 1.4*2x?(x^2+4y^2)
= x^2 + 2.8/x ?(x^6+174.24)
So, find where c'(x)=0
The algebra gets a bit tedious, but wolframalpha.com can help out there.
http://www.wolframalpha.com/input/?i=c+%3D+x%5E2+%2B+1.4*2x%E2%88%9A(x%5E2%2B4(6.6%2Fx%5E2)%5E2)
Imagine making a tent in the shape of a pyramid. Assume we want the volume to be 2.2m3 to sleep two or three people. Draw a picture identifying all appropriate variables. The floor of the tent is cheaper material than the rest: assume that the material making up the dome of the tent is 1.4 times as expensive per square meter than the material touching the ground.I would appreciate help with finding the equation to minimize the cost.
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