Monday

April 27, 2015

April 27, 2015

Posted by **Sarah** on Tuesday, April 8, 2008 at 6:57pm.

Use quadratic functions to solve.

- Algebra 2 -
**drwls**, Tuesday, April 8, 2008 at 7:51pmPut the points of the triangle at (0,0), (0,9) and (12,0)

Let the insribed rectangle have corners at:

(0,0), (x,0), (x,y'), and (0,y')

y' is on the hypotenuse of the triangle, so that

y = 12 - (4/3)x

So y' = 12 - (4/3)x

The rectangle area is

A(x) = x*y' = 12 x - (4/3)x^2

There is a maximum when dA/dx = 0

12 = (8/3) x

x = (3/8)*12 = 4.5

y' = 12 - (4/3)(9/2) = 5

The maximum rectangle area is xy'= 22 and the side lengths are 4.5 and 5.

- Algebra 2 -
**bobpursley**, Tuesday, April 8, 2008 at 7:55pmThis is normally a challenging calculus problem, however it can be worked with similar triangles, or algebra.

First, note the triangle is a right triangle. Lay it out so the 12 dimension is on the x axis, and the 9 is on the y axis.

Note the equation of the hypotenuse is

y=-9/12 x + 9

let one side of the rectangle going upward as h, and the side along the x axis as W. So the intersection with the hypotenuse is w,-9w/12 + 9

So the drill is to find the max area.

h= -9/12 w + 9

Area= hw= -9/12 w^2+9w

so when is Area max?

- Algebra 2 (correction) -
**drwls**, Tuesday, April 8, 2008 at 8:15pmI calculated the point on the hypotenuse incorrectly. It is

y' = 12 - (4/3)(9/2) = 6

The rectangle side lengths are 4.5 and 6, and the area is 27. That equals half the area of the triangle, (1/2)*9*12 = 54. BobPursley tells me that Euclid proved this in many different ways, without calculus.

Thanks to Bob for pointing this out

- Algebra 2 -
**david**, Monday, July 8, 2013 at 1:38pmA package with square ends has a combined length and girth (girth is the perimeter of a cross section) of 120 in. The surface area of the entire package is 3600 sq. in.

Determine the dimensions of thr package. s in.*s in.*l in.

one solution is: 11.08in*11.08in*75.68in.

find the other solution

hint: if 4S + length=120, then the length=120-4S

Round your answer to 2 decimals

Enter the 3 dimensions separated by comas