Math Help Please
1. "A straight section of railroad track crosses two highways at points that are 400m and 600m, respectively, from an intersection. Determine the dimensions of the largest rectangular lot that can be laid out in the triangle formed by the railroad and highways." [url=http://imageshack.us/photo/my-images/255/mathhj.jpg/]Pic to help[/url]
2. "A trough is made from a rectangular strip of metal 50cm wide. The metal is bent at right angles so it is "x" centimetres high along two sides. For what value of "x" is the cross-sectional area a maximum?" [url=http://imageshack.us/photo/my-images/684/mathy.jpg/] Pic to help[/url]
Thank you.
October 11, 2011
4 Comments • Newest first
[quote=Zyanid]1) Notice that you have two pairs of similar triangles. Choose one and set up a relationship that involves the the legs of the triangles.
I used the larger triangle (height of 400m and base of 600m) and the smaller triangle where the height (I assigned the variable y for this) is the also the height of the rectangle. I chose to assign the variable x for the base of the smaller triangle.
Set up the relationship: y/400 = x/600
This way you can solve for either x or y as needed.
Next you derive a formula for the area of the rectangle,which will help you get the dimensions.
A=(600-x)(y)
(600-x) comes from the fact that the length of the rectangle can't be the same length as the base of the larger triangle. Remember that the length of the rectangle + the base of the smaller triangle = the base of the larger triangle (600m).
Use the relationship we established earlier to solve for either x or y.
y/400 = x/600
y= (2x)/3
Substitute.
A = (600-x)(y)
A = (600-x)((2x)/3)
Differentiate:
A' = 400 - (4x)/3
0 = 400 - (4x)/3
x = 300 <-max (2nd Derivative Test)
Now solve for y:
y = (2x)/3
y = 200[/quote]
Thank you very much.
For 1 I think the biggest area is achieved when the space is a square. So you set the sides of the square to be x. The total area is 120000. Find the area of square, area of small triangle on top of square, and area of small triangle right of the square (all of these are of course in terms of x). Set this sum equal to 120000. Solve for x.
[quote=ohbabyitsanju]2) A = x(50-2x) = 50x - 2x^2
A' = 50 - 4x
0 = 50 - 4x
-50 = -4x
x = 12.5cm
EDIT: I failed the first time.[/quote]
k thanks.
2) A = x(50-2x) = 50x - 2x^2
A' = 50 - 4x
0 = 50 - 4x
-50 = -4x
x = 12.5cm
EDIT: I failed the first time.