Calc Help

[labron]

2 feet on each side of a square
to make an obtuse iscoceles triangle
with base 2 and altitude x from each
side of the square you are left
with a 4 pointed star which you fold to make
a pyramid. for what value x will the pyramid
have the maximum volume ?

this is a square with pointed triangles coming out of each side of the square

hint:
the volume of the of the pyramid is 1/3
the area of the base times the altitude.

 

[Deleted member 4993]

2 feet on each side of a square
to make an obtuse iscoceles triangle
with base 2 and altitude x from each
side of the square you are left
with a 4 pointed star which you fold to make
a pyramid. for what value x will the pyramid
have the maximum volume ?

this is a square with pointed triangles coming out of each side of the square

hint:
the volume of the of the pyramid is 1/3
the area of the base times the altitude.

Volume of a regular pyramid = 1/3 * area of the base * height

You can find the height of your pyramid by using Pythagorean theorem

Please share your work with us, indicating excatly where you are stuck - so that we may know where to begin to help you.

 

[labron]

yea i got what the volume is the 1/3bh

i just dont know where to get started

do i plug and chug? lol well like plug in till i find the best value of x or find the derevative of the fourmula?

 

[tkhunny]

You've already been instructed on where to start.

What is stopping you from using the deriative on Volume(b,h) = (1/3)*b*h?

Answer this question. Then ponder on the Pythagorean Theorem. It will solve this problem for you.

 

[labron]

i dont kno, okay so im stuck.

i did pygaotherom and the side of the square was 2 feet.

so 1sqrd + xsqrd = asqurd -> a is the sides of the triangles. All of them are same because isoceles

then i figured out the diagonal of the square (AB) which forms a 2 right triangles.

it was AB + 2x = 2 -> AB = 2 - 2x

then found one of the bases of the triangle AC -> BASE LINE OF TRIANGLE

A0 = 1/2(AB) = 1 - x The 0 stands for the center of the square.

ACsqurd = A02squrd + C0squrd = (1-x)sqrd + (1-x)sqrd

AC = [SQRD ROOT_(2)] x (1-x)

NOW trying to find the height

AE = 1/2(AC) = {[SQURDROOT_(2)] / 2 } x (1-x)


h squrd + AE squrd = a squrd

h squrd = a squrd - AE squrd

[ h squrd ]= [1] + [x squrd] - [2/4] x [(1-x)squrd]

i dont know if im doing it right at the end i got

h =[ [squr root 2] x ( 1+x ) ]/ 2