# SOLUTION: A rectangular piece of cardboard with perimeter 30 in., two parallel and equally spaced creases are made. The cardboard is then folded to make a prism with open ends that are equil

Algebra ->  -> SOLUTION: A rectangular piece of cardboard with perimeter 30 in., two parallel and equally spaced creases are made. The cardboard is then folded to make a prism with open ends that are equil      Log On

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 Click here to see ALL problems on Triangles Question 164928: A rectangular piece of cardboard with perimeter 30 in., two parallel and equally spaced creases are made. The cardboard is then folded to make a prism with open ends that are equilateral triangles. Find the volume of the prism as a function of x and what is the domain of V(x)? Thanks to those who can help!Answer by ankor@dixie-net.com(16524)   (Show Source): You can put this solution on YOUR website!A rectangular piece of cardboard with perimeter 30 in., two parallel and equally spaced creases are made. The cardboard is then folded to make a prism with open ends that are equilateral triangles. Find the volume of the prism as a function of x and what is the domain of V(x)? : Let x = the 3 equal distances, between the two folds then 3x = width of the cardboard and Let L = the length of cardboard ; Perimeter: 2(3x) + 2L = 30 Simplify divide by 2 3x + L = 15 L = (15-3x) : Find the height (h) of the equilateral triangle with sides = x h^2 + (x/2)^2 = x^2 h^2 = x^2 - (x/2)^2 h = : Vol of a triangular prism = *L*w*h In this problems L = (15-3x) w = x h = : V(x) = *(15-3x)*x* : You can tell by L, (15-3x), that x has to be less than 5 (the domain) : If you graph this: The max volume when x = 3.33, Domain 0 to <+5