Length, Largest Possible Area

[Guest]

Wire1.jpg (94.8 KiB) Viewed 1994 times

The diagram represents a wire 14" in length. The parts marked "x" are to be bent up to form figure CABD. Determine length of x so that the area within the U - shaped figure CABD is the largest possible.

Answer given with problem: 3 1/2 in.

How do I solve ?

[Guest]

The length of the horizontal bit of the U will be (14 - 2x) and the height of the U vertical arms will be "x"
So the area enclosed will be x(14 - 2x) this equals 14x - 2x^2. ..... So A = 14x - ax^2
This is a quadratic curve and an upside U in shape and its maximum will be when the rate of change of A with respect to "x" is Zero (a horizontal tangent to the curve)
you can plot the curve on graph paper and read the "x" value where the curve is maximum to get the answer.

To do it mathematically you have to differentiate A with respect to x ...that is find dA/dx (read as "d" "A" by "d" "x") and put it equal to Zero for maximum point.

So ..... dA/dx of (14x - 2x^2 ) = 14 - 4x

Put it equal to Zero ......14 - 4x = 0

Re-arrange .... 4x = 14
So ....... x = 3.5 or 3 1/2 as in your answer.

[Guest]

So the area enclosed will be x(14 - 2x) this equals 14x - 2x^2. ..... So A = 14x - ax^2

Sorry ....... this bit should read......as below

So the area enclosed will be x(14 - 2x) this equals 14x - 2x^2. ..... So A = 14x - 2x^2

[Guest]

This is a quadratic curve and an upside U in shape

and sorry this bit should read ....as below.....

This is a quadratic curve and an upside down U in shape

The curve will be an inverted U ................ the centre of the curve at the top

[Guest]

' So the area enclosed will be x(14 - 2x) this equals 14x - 2x^2. ..... So A = 14x - 2x^2

To do it mathematically you have to differentiate A with respect to x ...that is find dA/dx (read as "d" "A" by "d" "x") and put it equal to Zero for maximum point.

So ..... dA/dx of (14x - 2x^2 ) = 14 - 4x "


I don't understand these parts.

[Guest]

You would need to look up the basics of finding rates of change of functions. That is finding the slope of straight line graphs and functions.
For quadratic function you find the slope of the tangent to the curve at the required point. OR in this case find the value of "x" when the slope of the tangent to the curve is zero

[Guest]

I have no clue what that is.

[Guest]

Did you plot the graph of A versus x. Plot A on the Y axis against x on the X axis.

You should see the graph rising up and turning and coming back down again. At the top a tangent to the curve will be horizontal, this is the point of maximum area and you can read off the value for x.

[Guest]

I don't know how.

[Guest]

Make a table and work out the area for each value of X from 0.0 to 5.0 in 0.5 steps
and plot the points on a graph paper, join the dots with a line to form the curve

[Guest]

Although differentiation to find the maximum is the obvious way to solve this question, it is not actually needed. You can solve it by simply completing the square
http://www.mathcentre.ac.uk/resources/u ... 2009-1.pdf

Area [tex]= 14x-2x^2[/tex]
[tex]= 24.5-2(x-3.5)^2[/tex]
Squares are always non-negative so to maximize the area make the square term [tex]0[/tex], which happens when [tex]x=3.5[/tex].

Hope this helped,

R. Baber.

[Guest]

Ok, I understand completing the square is another method but I always found this method was more more complicated.
Graphic solutions give an answer as accurate as you can read from the graph, then trial and error will give a more exact solution.
So one can read 3,5 from the graph and then test it in the expression.

[Guest]

Thanks, R. Baber, that helped.

[Guest]

Guest,

How did you do the graph ?

[Guest]

I did it using excel spreadsheet, but it is better for you if you draw it yourself manually.
Create a table as shown on left of the graph .....it is the value of the expression 14X - 2xX^2 for each value of X.
In column 1 put in the values of X, start at 0 and go up to 7 in 0.5 steps
For column 2 multiply each X value by 14 and put it in column 2
For column 3 multiply each X value by 2 time X value squared and put it in column 3
For column 4 subtract each column 3 value from each corresponding column 2 value and put in in column 4
Plot the values of column 1 as X value and corresponding column 4 value as Y value.
You should get the same shape of graph and you will see that the maximum area is when X = 3.5 and the corresponding Y value for the area is 24.5

If you draw a tangent to the curve you will see it is horizontal at X = 3.5 and you will see the Y area value is 24.5.

[Guest]

Ok, thanks Guest.