math question! please help

Question

1) the perimeter of a right triangle is 24in. find the dimnsions of the triangle if the sides are all whole number lengths 

 

2) what size corners should be cut from a piece of cardboard that measures 30in. by 30in. to make an open-top box with the greatest possible volume? 

Answer 1

1) Think of all sets of three whole numbers which add to 24, and note that a right triangle has a hypotenuse which is greater than the two legs, yet smaller than their sum (thus the hypotenuse cannot be 12 or greater, and therefore no three numbers can be 12 or greater). Then check the numbers with pythagorean's theorem

11 + 6 + 7

-> check: 121 = 36 + 49? - Nope, hypotenuse is too large

10 + 7 + 7

-> check: 100 = 49 + 49? - Nope, but close

10 + 6 + 8

-> check: 100 = 36 + 64? - Yes!

 

Thus, the dimensions of the right triangle are:

10 (hypotenuse), 6, 8

 

2) Maximum volume of a rectangular prism is a cube, thus we have a design from a piece of cardboard which would have the base in the middle as a square, and 4 flaps extending to the edge of the cardboard. The sum of three edges (the left flap to right flap) would be 30in (Note that the cardboard is a square). This length involves three edges lined up, thus 3*l = 30. So the length, height and width of the box is 10 inches. This requires cutting off 4 squares or corners of the cardboard. The size of these squares will end up being 100 sq inches for each corner cut off (10^2). Thus 100 square inches should be cut off at each corner in the shape of a square.