SOLUTION: Hey, I just need help with two word problems. 1st one: A lot is in the shape of a right triangle. The shorter leg measures 180 m. The hypotenuse is 60 m longer than the length o

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Question 166560: Hey, I just need help with two word problems.
1st one: A lot is in the shape of a right triangle. The shorter leg measures 180 m. The hypotenuse is 60 m longer than the length of the longer leg. How long is the longer leg??
2nd one: A rectangular piece of cardboard measuring 28 inches by 29 inches is to be made into a box with an open top by cutting equal size squares from each corner and folding up the sides. Let x represent the length of a side of each such square. For what value of x will the volume be a maximum? If necessary, round to 2 decimal places.
Thank u
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
1st one: A lot is in the shape of a right triangle. The shorter leg measures 180 m. The hypotenuse is 60 m longer than the length of the longer leg. How long is the longer leg??
.
Let x = length of longer leg
then
x+60 = hypotenuse
.
Applying pythagorean's theorem:
180^2 + x^2 = (x+60)^2
180^2 + x^2 = (x+60)^2
32400 + x^2 = x^2 + 120x + 3600
32400 = 120x + 3600
28800 = 120x
240 meters = x (longer leg)
.
2nd one: A rectangular piece of cardboard measuring 28 inches by 29 inches is to be made into a box with an open top by cutting equal size squares from each corner and folding up the sides. Let x represent the length of a side of each such square. For what value of x will the volume be a maximum? If necessary, round to 2 decimal places.
.
Volume = x(28-2x)(29-2x)
Volume = x(812-56x-58x+4x^2)
Volume = x(812-114x+4x^2)
Volume = 812x-114x^2+4x^3
Volume = 4x^3-114x^2+812x
.
Taking the derivative:
4x^3-114x^2+812x
12x^2-228x+812 = 0
6x^2-114x+406 = 0
Solving the above with the quadratic equation we get:
x = {14.25, 4.75}
.
Solution: 4.75 inches
.
Details of quadratic equation:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 6x%5E2%2B-114x%2B406+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-114%29%5E2-4%2A6%2A406=3252.

Discriminant d=3252 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--114%2B-sqrt%28+3252+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-114%29%2Bsqrt%28+3252+%29%29%2F2%5C6+=+14.2521924764611
x%5B2%5D+=+%28-%28-114%29-sqrt%28+3252+%29%29%2F2%5C6+=+4.74780752353892

Quadratic expression 6x%5E2%2B-114x%2B406 can be factored:
6x%5E2%2B-114x%2B406+=+6%28x-14.2521924764611%29%2A%28x-4.74780752353892%29
Again, the answer is: 14.2521924764611, 4.74780752353892. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+6%2Ax%5E2%2B-114%2Ax%2B406+%29