Another max min: A rectangle is bounded by the positive x-axis, the positive y-axis, and y = mx + c.

[pazzy78]

So, I have been trying to get the area of the rectangle in terms of m and c, but trying to rewrite the h and l of the rectangle just gives me circular results, no help .. I decided to use similar triangles
to get them expressed in different terms.

Maybe this is the wrong direction...

 

[stapel]

So, I have been trying to get the area of the rectangle in terms of m and c, but trying to rewrite the h and l of the rectangle just gives me circular results, no help .. I decided to use similar triangles
to get them expressed in different terms.

Maybe this is the wrong direction...

(a) The rectangle will have some width [imath]x[/imath]. At that width, what is the expression for the height of the rectangle? (Hint: Use the line equation.)

(b) The area is found by multiplying the width by the height. What expression does this give you, for the area [imath]A[/imath] in terms of [imath]x[/imath], [imath]c[/imath], and [imath]m[/imath]? (Hint: Multiply)

(c) Taking the derivative and setting it equal to zero, what expression do you get for [imath]x[/imath] in terms of [imath]c[/imath] and [imath]m[/imath]?

 

[pazzy78]

Nice, the right answer is c^2/4m but I get - c^2/4m .

I don't like getting a negative area, but in this case do we just take the absolute value like when integrating under curves that are in the negative zone of a graph...

 

[pka]

Looking at the diagram the lower right corner of the square is [imath]\left(x_0,0\right)[/imath].
The height of the square is [imath]m\;x_0+c[/imath]
Therefore the area of the square is [imath]A=x_0\left(m\;x_0+c\right)[/imath].
What do we do to maximize the area of the square?

 

[topsquark]

Nice, the right answer is c^2/4m but I get - c^2/4m .

I don't like getting a negative area, but in this case do we just take the absolute value like when integrating under curves that are in the negative zone of a graph...

Your answer is correct. Note that your slope, m, is actually negative, so your area is positive, as it should be.

Either your line equation was meant to be y =-mx + c, or your solution key should have said [imath]-c^2/(4m)[/imath] (note the parentheses!) One way or another, there was a typo.

-Dan

 

[Steven G]

I would have phrased what @stapel said differently.
Pick some arbitrary point, (x,y), on the line. This point on the line makes a unique triangle. Now continue as you already did.