An open box contains 80cm^3 and is made from a square piece of tinplate with 3cm squares cut from each of its 4 corners. Find the dimensions of the original piece of tinplate.

Question

anonymous · Answer

the height of the box will be 3cm and the length and width are both (x-6) where x is the original dimension of the tinplate 
 3(x-6)(x-6)=80 
 3(x²-12x+36)=80 
 3x²-12x+108=80 
 3x²-12x+28=0 
 use the quadratic formula 
 x=11.16 -0.84 ignore -0.84 because you cannot have a negative dimension 
 so the original dimensions are 11.16cm*11.16cm