SOLUTION: It is desired to have an open bin with a square bottom,rectangular sides,and a height of 3m.If the material for the bottom costs $5.40 per square meter and the material for the sid

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: It is desired to have an open bin with a square bottom,rectangular sides,and a height of 3m.If the material for the bottom costs $5.40 per square meter and the material for the sid      Log On


   



Question 783865: It is desired to have an open bin with a square bottom,rectangular sides,and a height of 3m.If the material for the bottom costs $5.40 per square meter and the material for the sides costs $2.40 per square meter and the material for the sides costs $2.40 per square meter,what is the volume of the bin that can be constructed for $63 worth of material?
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
One bottom, and four sides.

Let x = length of side of bottom, a square shape.
Let 3 = length of side for dimension of the three-D figure, perpendicular to the square bottom, which is the height of the bin. 3 is how tall the bin.


Area of the bottom of the bin is x%5E2, and the cost is 5.40%2Ax%5E2.
Area for all four sides of the bin is 4%2A%283x%29, and the cost is 2.40%2A%284%29%283x%29.

Wanted is cost be $63.
5.40x%5E2%2B%282.40%294%283x%29=63
5.4x%5E2%2B2.4%2A12x=63
5.4x%5E2%2B28.8x=63
Multiply by 10 and divide by 6,
9x%5E2%2B48x=70, which is maybe the easiest equivalent of the equation to work with.
Put in more general form:
highlight%289x%5E2%2B48x-70=0%29

Directly use general solution to quadratic equation.
Discrim, %2848%29%5E2-4%2A9%2A%28-70%29=48%5E2%2B2520=4824, which is 4%2A1201
x=%28-48%2Bsqrt%284%2A1201%29%29%2F%282%2A9%29
x=%28-48%2B2%2Asqrt%281201%29%29%2F%282%2A9%29
highlight%28x=%28-24%2Bsqrt%281201%29%29%2F9%29