Optimization Problem

[hcinic]

Construct a rectangle by placing one side on the line y=10-2x, one vertex on the positive x-axis, and one vertex on the positive y-axis. What are the dimensions of the rectangle with the max area when constructed this way?

Well, you end up with 3 triangles around the rectangle, but I don't see how maximizing or minimizing these will help. Correct me if i'm wrong, but since the side of the rectangle is on the line y=10-2x you are not able to solve like the easier question of one SIDE on positive x-axis and one SIDE on positive y-axis and one VERTEX on line y=10-2x which would lead to maximizing the derivative of this equation A(x)=x(10-2x)...any thoughts? Although I still think the maximum of both of these rectangles constructed inside these constraints are the same size...

 

[mmm4444bot]

You could come up with a function that inputs the distance from the origin to the vertex on the positive x-axis (that's your x-value) and outputs the rectangle's area.

In this regard, the optimization is found using the same overall approach as your earlier exercise.

It's going to take several steps to build the area function.

One side of the rectangle lies on the given line (in Quad I). Hence, the opposite side of the rectangle must lie on a line that is parallel to the given line.

Draw a diagram, and sketch a representative rectangle.

Using x as the x-coordinate of the vertex on the x-axis, consider how you would determine the distance between the two parallel lines.

That distance is one dimension of the rectangle. The other dimension should be obvious, from your diagram (it's the distance from the y-intercept to the x-intercept, along the variable parallel line).

 

[hcinic]

Yes, i see that the parallel lines have the same slope and the line perpendicular to y=-2x+10 has the slope of the negative recipricol: 1/2...thinking out loud...

 

[lookagain]

\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \)\(\displaystyle \text{B}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ *\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ * \ \ \ \ \ \ \ *\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ * \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ * \)
\(\displaystyle \ \ \ \ \ \ \ \ \ * \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ *\)
\(\displaystyle \ \ \ \ \ \ \ \ ***************\)
\(\displaystyle \ \ \ \ \ \)\(\displaystyle \text{A}\)\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \) \(\displaystyle \text{C}\)



You can also orient the triangle on the xy-axes with point A at the origin (0, 0), and points B and C
as pictured above. Angle B is a right angle. Side AC is a hypotenuse. The length of AB = 5,
and the length of BC = 10.


Suggestions:

i) Work out the length of AC using the Pythagorean Theorem.


ii) Get the coordinates of point C from that by inspection.


iii) Work out the coordinates of point B using similar right triangles.


iv) Pick two variable x-values, \(\displaystyle x_1 \ and \ x_2,\) on line segment AC, with \(\displaystyle x_1 < x_2.\)


v) Work out the equation of the line that AB lies on, as well as work out the equation
of the line that BC lies on.


vi) Work out a function for the height in terms of \(\displaystyle x_1.\)


vii) Set that height function equal to the linear function for BC, and solve for a \(\displaystyle "new" \ value \ \ of \ \ x_1.\)
\(\displaystyle (x_2 \ \ equals \ \ this \ \ "new" \ \ value \ \ of \ \ x_1.)\)


viii) Work out the function for the base of the rectangle in terms of \(\displaystyle x_1. \ \)

You can use \(\displaystyle \ x_2 - x_1 \ \ for \ \ that.\)


ix) Multiply the function of the height by the function of the base.


x) Take the derivative of that product, set it equal to zero, and solve for the original/"old" \(\displaystyle \ x_1.\)

 

[mmm4444bot]

you are not able to solve like the easier question of one SIDE on positive x-axis and one SIDE on positive y-axis and one VERTEX on line y=10-2x which would lead to maximizing the derivative of this equation A(x)=x(10-2x)...any thoughts?

Did you finish this exercise?

I just worked it out, using my previously-posted approach (with similar triangles giving the distance between parallel lines), and I wound up with the same function as your A(x) above, heh, heh, heh.