Optimization - rectangle inscribed in a right triangle

In summary, the dimensions of the rectangle with the greatest possible area inscribed in a right triangle with sides 3 cm, 4 cm, and 5 cm are l = 1.2 cm and w = 2.5 cm, with a maximum area of 3 cm^2. The area can be expressed as A(y) = 3y - (3/4)y^2, where y is the length of the rectangle. It can also be found by setting the proportions l/(4-y) = 3/5 and w/y = 5/4.
  • #1
ghostanime2001
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0

Homework Statement


A rectangle is to be inscribed in a right triangle having sides 3 cm, 4 cm and 5 cm, as shown on the diagram. Find the dimensions of the rectangle with greatest possible area.

Homework Equations


1. [itex]x^{2}+y^{2}=w^{2}[/itex] in terms of [itex]w=\sqrt{x^{2}+y^{2}}[/itex]

2. [itex]\dfrac{x}{3}=\dfrac{y}{4}[/itex]

3. [itex]\dfrac{l}{3}=\dfrac{4-y}{5}[/itex]

4. [itex]\dfrac{l}{x}=\dfrac{4-y}{w}[/itex]

The Attempt at a Solution



Area of the rectangle is [itex]A=lw[/itex]. But, we want an expression for area in terms of one variable.

rewriting equation 2 for [itex]x[/itex] we can get an expression in [itex]y[/itex]

[itex]4x=3y[/itex]

[itex]x=\dfrac{3y}{4}[/itex]

which we can substitute in equation 1 to find width [itex]w[/itex] in terms of [itex]y[/itex]

[itex]w=\sqrt{x^{2}+y^{2}}[/itex]

[itex]w=\sqrt{\left(\dfrac{3y}{4}\right)^{2}+y^{2}}[/itex]

[itex]w=\sqrt{\dfrac{9}{16}y^{2}+y^{2}}[/itex]

[itex]w=\sqrt{\dfrac{25}{16}y^{2}}[/itex]

[itex]w=\dfrac{5}{4}y[/itex]

Using similar triangles of the upper right, smaller right triangle and the bigger, outer triangle we set a relationship between the base of smaller triangle and the base of the outer triangle, and, the hypotenuse of the smaller triangle and the hypotenuse of the outer triangle. We solve for the length of the rectangle in terms of [itex]y[/itex] using equation 3.

[itex]\dfrac{l}{3}=\dfrac{4-y}{5}[/itex]

[itex]5l=3\left(4-y\right)[/itex]

[itex]l=\dfrac{3\left(4-y\right)}{5}[/itex]

We now have the length [itex]l[/itex] and width [itex]w[/itex] in terms of one variable [itex]y[/itex]. We substitute the expressions in the area of the rectangle to find the area in one variable [itex]y[/itex]

[itex]A\left(y\right)=l\left(y\right)w\left(y\right)[/itex]

[itex]A\left(y\right)=\dfrac{3\left(4-y\right)}{5}\dfrac{5}{4}y[/itex]

[itex]A\left(y\right)=\dfrac{15y\left(4-y\right)}{20}[/itex]

[itex]A\left(y\right)=\dfrac{3y\left(4-y\right)}{4}[/itex]

[itex]A\left(y\right)=\dfrac{12y-3y^{2}}{4}[/itex]

[itex]A\left(y\right)=3y-\dfrac{3}{4}y^{2}[/itex]

Differentiating this expression and setting [itex]A'\left(y\right)=0[/itex] gives [itex]0=3-\dfrac{3}{2}y[/itex]

solving for [itex]y[/itex] gives

[itex]\dfrac{3}{2}y=3[/itex]

[itex]y=2[/itex]

substitute [itex]y=2[/itex] in the length [itex]l=\dfrac{3\left(4-y\right)}{5}[/itex] and width [itex]w=\dfrac{5}{4}y[/itex] equations respectively,

[itex]l=\dfrac{3\left(4-\left(2\right)\right)}{5}=\dfrac{6}{5}[/itex]

[itex]w=\dfrac{5}{4}\left(2\right)=\dfrac{5}{2}[/itex]

Therefore, the rectangle with dimensions [itex]l=\dfrac{6}{5}=\text{1.2 cm}[/itex] and [itex]w=\dfrac{5}{2}=\text{2.5 cm}[/itex] has the maximum possible area of [itex]\text{3 }cm^{2}[/itex]

*above I used the more tedious way I guess...

My question is if I can do it using equation 4 since [itex]l[/itex] and [itex]w[/itex] are already there and cross multiplying would give me area automatically. I would have an expression in [itex]x[/itex] and [itex]y[/itex] but I can use equation 2 to find either [itex]x[/itex] or [itex]y[/itex] and set up the area in one variable. Would this way be the more correct method or give the same answer as above? Thanks !
 

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  • #2
ghostanime2001 said:
My question is if I can do it using equation 4 since [itex]l[/itex] and [itex]w[/itex] are already there and cross multiplying would give me area automatically. I would have an expression in [itex]x[/itex] and [itex]y[/itex] but I can use equation 2 to find either [itex]x[/itex] or [itex]y[/itex] and set up the area in one variable. Would this way be the more correct method or give the same answer as above? Thanks !

That works! I get an answer of 3. Did you try it out and see how the answers compared? You can use the proportion ##\frac{x}{3} = \frac{y}{4}## to get the equation in terms of only x or y and then take the derivative, etc.

As for more "correct," every method that arrives at a solution is correct, but this way might be more elegant because it only takes 5 lines or so of equations.
 
  • #3
ghostanime2001 said:

Homework Statement


A rectangle is to be inscribed in a right triangle having sides 3 cm, 4 cm and 5 cm, as shown on the diagram. Find the dimensions of the rectangle with greatest possible area.

Homework Equations


1. [itex]x^{2}+y^{2}=w^{2}[/itex] in terms of [itex]w=\sqrt{x^{2}+y^{2}}[/itex]

2. [itex]\dfrac{x}{3}=\dfrac{y}{4}[/itex]

3. [itex]\dfrac{l}{3}=\dfrac{4-y}{5}[/itex]

4. [itex]\dfrac{l}{x}=\dfrac{4-y}{w}[/itex]

The Attempt at a Solution



Area of the rectangle is [itex]A=lw[/itex]. But, we want an expression for area in terms of one variable.

rewriting equation 2 for [itex]x[/itex] we can get an expression in [itex]y[/itex]

[itex]4x=3y[/itex]

[itex]x=\dfrac{3y}{4}[/itex]

which we can substitute in equation 1 to find width [itex]w[/itex] in terms of [itex]y[/itex]

[itex]w=\sqrt{x^{2}+y^{2}}[/itex]

[itex]w=\sqrt{\left(\dfrac{3y}{4}\right)^{2}+y^{2}}[/itex]

[itex]w=\sqrt{\dfrac{9}{16}y^{2}+y^{2}}[/itex]

[itex]w=\sqrt{\dfrac{25}{16}y^{2}}[/itex]

[itex]w=\dfrac{5}{4}y[/itex]

Using similar triangles of the upper right, smaller right triangle and the bigger, outer triangle we set a relationship between the base of smaller triangle and the base of the outer triangle, and, the hypotenuse of the smaller triangle and the hypotenuse of the outer triangle. We solve for the length of the rectangle in terms of [itex]y[/itex] using equation 3.

[itex]\dfrac{l}{3}=\dfrac{4-y}{5}[/itex]

[itex]5l=3\left(4-y\right)[/itex]

[itex]l=\dfrac{3\left(4-y\right)}{5}[/itex]

We now have the length [itex]l[/itex] and width [itex]w[/itex] in terms of one variable [itex]y[/itex]. We substitute the expressions in the area of the rectangle to find the area in one variable [itex]y[/itex]

[itex]A\left(y\right)=l\left(y\right)w\left(y\right)[/itex]

[itex]A\left(y\right)=\dfrac{3\left(4-y\right)}{5}\dfrac{5}{4}y[/itex]

[itex]A\left(y\right)=\dfrac{15y\left(4-y\right)}{20}[/itex]

[itex]A\left(y\right)=\dfrac{3y\left(4-y\right)}{4}[/itex]

[itex]A\left(y\right)=\dfrac{12y-3y^{2}}{4}[/itex]

[itex]A\left(y\right)=3y-\dfrac{3}{4}y^{2}[/itex]

Differentiating this expression and setting [itex]A'\left(y\right)=0[/itex] gives [itex]0=3-\dfrac{3}{2}y[/itex]

solving for [itex]y[/itex] gives

[itex]\dfrac{3}{2}y=3[/itex]

[itex]y=2[/itex]

substitute [itex]y=2[/itex] in the length [itex]l=\dfrac{3\left(4-y\right)}{5}[/itex] and width [itex]w=\dfrac{5}{4}y[/itex] equations respectively,

[itex]l=\dfrac{3\left(4-\left(2\right)\right)}{5}=\dfrac{6}{5}[/itex]

[itex]w=\dfrac{5}{4}\left(2\right)=\dfrac{5}{2}[/itex]

Therefore, the rectangle with dimensions [itex]l=\dfrac{6}{5}=\text{1.2 cm}[/itex] and [itex]w=\dfrac{5}{2}=\text{2.5 cm}[/itex] has the maximum possible area of [itex]\text{3 }cm^{2}[/itex]

*above I used the more tedious way I guess...

My question is if I can do it using equation 4 since [itex]l[/itex] and [itex]w[/itex] are already there and cross multiplying would give me area automatically. I would have an expression in [itex]x[/itex] and [itex]y[/itex] but I can use equation 2 to find either [itex]x[/itex] or [itex]y[/itex] and set up the area in one variable. Would this way be the more correct method or give the same answer as above? Thanks !

It is a lot easier to use
[tex]\frac{l}{4-y} = \frac{3}{5} \Longrightarrow l = (3/5)(4-y)\\
\frac{w}{y} = \frac{5}{4} \Longrightarrow w = (5/4)y[/tex]
giving ##A = l w = (3/4) y(4-y) = 3 y - (3/4)y^2.##
 
  • #4
Ray Vickson, I had overlooked this relation! Your relation directly solves [itex]w[/itex] in terms of [itex]y[/itex] avoiding going through using pythagorean theorum and equation 2. Am I right ? or am I right ?

The first relation is the same one I acquired. It is just written differently.

[itex]\dfrac{w}{y}=\dfrac{5}{4}\Rightarrow w=\dfrac{5}{4}y[/itex]

scurty, I have used equation 4 and equation 2 to solve for area [itex]lw[/itex] in terms of [itex]x[/itex] (instead of [itex]y[/itex] as previously used) and I get

[itex]A=lw=4x-\dfrac{4}{3}x^{2}[/itex]

compared to [itex]A=lw=3y-\dfrac{3}{4}y^{2}[/itex]
 
  • #5
ghostanime2001 said:
scurty, I have used equation 4 and equation 2 to solve for area [itex]lw[/itex] in terms of [itex]x[/itex] (instead of [itex]y[/itex] as previously used) and I get

[itex]A=lw=4x-\dfrac{4}{3}x^{2}[/itex]

compared to [itex]A=lw=3y-\dfrac{3}{4}y^{2}[/itex]

I can't tell if you are just acknowledging that the method worked (if so, good job!) or you are confused. Sorry, haha, could you just clear that up for me?
 
  • #6
the area equation in terms of [itex]x[/itex] is correct, no ? I just want to know if I did my algebra right. I substituted the value of [itex]x=\dfrac{3}{2}=\text{1.5 cm}[/itex] into the area equation and it comes out to [itex]\text{3 }cm^{2}[/itex]. Do you have the same numbers & answer ?
 
  • #7
Yes, it's correct! There's more than one way to arrive at the solution.
 
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Related to Optimization - rectangle inscribed in a right triangle

1. How do you find the maximum area of a rectangle inscribed in a right triangle?

In order to find the maximum area of a rectangle inscribed in a right triangle, you need to understand the properties of right triangles. The maximum area will always be achieved when the rectangle's vertices are placed on the hypotenuse of the right triangle, dividing it into three smaller triangles with equal areas.

2. What is the formula for finding the area of a rectangle inscribed in a right triangle?

The formula for finding the area of a rectangle inscribed in a right triangle is A = 1/2 * b * h, where b is the length of the base of the right triangle and h is the height of the right triangle. The base and height of the rectangle will also be the base and height of the smaller triangles formed by the rectangle on the hypotenuse.

3. How do you determine the dimensions of the rectangle with the maximum area?

To determine the dimensions of the rectangle with the maximum area, you can use the Pythagorean theorem to find the length of the rectangle's sides. The sides will be equal to the base and height of the smaller triangles formed by the rectangle on the hypotenuse. The dimensions of the rectangle will then be the base and height of the smaller triangles multiplied by 2.

4. Can a rectangle be inscribed in any right triangle?

Yes, a rectangle can be inscribed in any right triangle as long as the rectangle's sides are not longer than the shorter sides of the right triangle. In other words, the rectangle must fit within the triangle without extending beyond the triangle's boundaries.

5. What is the significance of finding the maximum area of a rectangle inscribed in a right triangle?

Finding the maximum area of a rectangle inscribed in a right triangle is significant because it allows for efficient use of space. This concept is often used in optimization problems in mathematics and real-world applications, such as maximizing the area of a garden or finding the most cost-effective dimensions for a product's packaging.

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