Maximizing Area: Solving Quadratic Applications with Triangles and Rectangles

[hackedagainanda]

Homework Statement

The front face of a house is in the shape of a rectangle with a Queen post roof truss above. The length of the rectangular region is 3 times the height of the truss. The height of the rectangle is 2 ft more than the height of the truss. If the total area of the front of the house is 336 ft^2 determine the length and the width of the rectangular region

Relevant Equations

Area for rectangle: A = LW
Area for a triangle (1/2)bh

So the top of the structure is a triangle with height x. and the height of the rectangle is 2 + x, and the length is 3x.

I'm unsure where to go from here. I tried using the formula and getting 3x^2 + 6x +x = 3x^2 + 7x -336 =0 I applied the quadratic formula but it gave me non-integer solutions. I may be reading the problem wrong or not getting the right equation from the prompt.

 

[FactChecker]

1) Why do you think the answer needs to be an integer?
EDIT: The following is incorrect. The area given is for the entire side, including the triangle on top.
2) Why are you adding an extra x? All the information is about the length and width of a rectangle and the question asks about the area of the rectangle. I don't see the need for adding another x.
3x(x+2) = 3x^2+6x=336.

 

[sysprog]

Hmm, @hackedagainanda, isn't your anwer numerically the same as that of @FactChecker? Doesn't 'whatever-336=0' mean the same thing as 'whatever=336'?

 

[Mark44]

Homework Statement:: The front face of a house is in the shape of a rectangle with a Queen post roof truss above. The length of the rectangular region is 3 times the height of the truss. The height of the rectangle is 2 ft more than the height of the truss. If the total area of the front of the house is 336 ft^2 determine the length and the width of the rectangular region
Relevant Equations:: Area for rectangle: A = LW
Area for a triangle (1/2)bh

3x^2 + 6x +x = 3x^2 + 7x -336 =0

The business of "Queen post roof truss" is a red herring. The only thing important about the truss is its height, which we can call x. The height of the Queen posts (I had to look up "queen post truss") is irrelevant in this problem.

The equation in the OP does not represent the sum of the rectangular area and the area of the roof truss.
"If the total area of the front of the house is 336 ft^2..."

In my formulation of the problem, there are two integer solutions, one of which is negative.

3x(x+2) = 3x^2+6x=336.

This equation is incorrect also, as it doesn't include the area of the truss.

Doesn't 'whatever-336=0' mean the same thing as 'whatever=336'?

Yes, but the OP and @FactChecker don't have the same equations.

 

[sysprog]

Yes, but the OP and @FactChecker don't have the same equations.

Right, of course, but can't 'whatever' [represent / be substituted for] the lhs of either equation?

 

[FactChecker]

This equation is incorrect also, as it doesn't include the area of the truss.

Yes, but the OP and @FactChecker don't have the same equations.

Sorry. I stand corrected. I thought that the area of 336 was for the rectangle, but it is not. It is for the entire side, including the triangle on the top.

 

[Mark44]

Right, of course, but can't 'whatever' [represent / be substituted for] the lhs of either equation?

Sure, and this is what I agreed with you on.

 

[hackedagainanda]

1) Why do you think the answer needs to be an integer?

The other questions in this section usually say round to the nearest tenth, hundreths, etc. when its a non-integer.

The business of "Queen post roof truss" is a red herring. The only thing important about the truss is its height, which we can call x. The height of the Queen posts (I had to look up "queen post truss") is irrelevant in this problem.
View attachment 289182
The equation in the OP does not represent the sum of the rectangular area and the area of the roof truss.
"If the total area of the front of the house is 336 ft^2..."

In my formulation of the problem, there are two integer solutions, one of which is negative.

Can I get a hint at formulating the problem?

 

[Mark44]

Can I get a hint at formulating the problem?

Look at the drawing I posted. In my drawing, x is the height of the truss. what are the dimensions of the rectangular part of the front of the house.

Again, the problem statement says this: "the total area of the front of the house is 336 ft^2..."
The front of the house includes the rectangular part and the area defined by the truss.

 

[hackedagainanda]

The dimensions are 3x(x +2) and the truss is 2 triangles so its x(3x) so its 3x^2 + 3x^2 +6x = 336

Write in standard form and get 6x^2 +6x - 336 = 0 The quadratic formula gets me x = 7, and x = -8

 

[Mark44]

The dimensions are 3x(x +2) and the truss is 2 triangles so its x(3x) so its 3x^2 + 3x^2 +6x = 336

Write in standard form and get 6x^2 +6x - 336 = 0 The quadratic formula gets me x = 7, and x = -8

Obviously, x = -8 can't be a solution. Does the solution x = 7 produce the right area for the front of the house? You should always check your solution for these types of problems.

 

[hackedagainanda]

It does check 3x 7 x (7 +2) = 21 x 9 = 189, + 3(7^2)= 147. 147 + 189 = 336. So the area of the rectangle is 189 ft^2.Thanks for the help, I'll have to remember draw diagrams on the test.

 

[SammyS]

It does check 3x 7 x (7 +2) = 21 x 9 = 189, + 3(7^2)= 147. 147 + 189 = 336. So the area of the rectangle is 189 ft^2.Thanks for the help, I'll have to remember draw diagrams on the test.

What's the combined area of your two triangles?

 

[hackedagainanda]

What's the combined area of your two triangles?

147 ft^2

 

[Mark44]

I'll have to remember draw diagrams on the test.

It's a good idea to draw a diagram for these kinds of problems, whether it's a test or not.

 

[SammyS]

147 ft^2

So then each triangle has an area of 73.5 ft².

But with two triangles sitting on top of the rectangle, each would have a base of 21/2 = 10.5 ft and altitude of 7 ft. That gives an area of 73.5/2 =36.75 ft² for each and a total are of 73.5 ft² for the two triangles combined.

That does not add up to the required total area of 336 ft².

 

[hackedagainanda]

So then each triangle has an area of 73.5 ft².

But with two triangles sitting on top of the rectangle, each would have a base of 21/2 = 10.5 ft and altitude of 7 ft. That gives an area of 73.5/2 =36.75 ft² for each and a total are of 73.5 ft² for the two triangles combined.

That does not add up to the required total area of 336 ft².

The 189 ft^2 is correct though... Which is the correct answer, but did I get the right answer for the wrong reason?

 

[Mark44]

The 189 ft^2 is correct though... Which is the correct answer

I don't think so. If 189 sq. ft. is the posted answer, whoever solved the problem is working a different one. I get an area for the rectangle part of the house front of 240 sq. ft.

 

[hackedagainanda]

Is the quadratic 6x^2 + 6x - 336 = 0 the wrong equation?

 

[SammyS]

Is the quadratic 6x^2 + 6x - 336 = 0 the wrong equation?

Yes, it's wrong.

The dimensions are 3x(x +2) and the truss is 2 triangles so its x(3x) so its 3x^2 + 3x^2 +6x = 336

Write in standard form and get 6x^2 +6x - 336 = 0 The quadratic formula gets me x = 7, and x = -8

You're not very clear regarding the base and height for the triangles.

Start with an individual triangle. What is the expression for the base? What is the expression for the height?

 

[haruspex]

the truss is 2 triangles so its x(3x)

The width of each triangle being...?

 

[hackedagainanda]

(1/2)x(1.5x) So I get 2(1/2)x(1.5x) so adding the area of the rectangle I get 3x^2 + 6x + 1.5x for a total of
4.5x^2 +6x - 336 = 0. The quadratic formula gets me 8, and - 28/3. The area of each triangle is (1/2)8 x 12 which is 48. The area of each triangle is 48 ft^2 both combined is 96 ft^2 and 96 ft^2 + 240 ft^2 = 336 ft ^2 so the answer checks.

 

[Mark44]

(1/2)x(1.5x) So I get 2(1/2)x(1.5x) so adding the area of the rectangle I get 3x^2 + 6x + 1.5x for a total of
4.5x +6x - 336 = 0. The quadratic formula gets me 8, and - 28/3. The area of each triangle is (1/2)8 x 12 which is 48. The area of each triangle is 48 ft^2 both combined is 96 ft^2 and 96 ft^2 + 240 ft^2 = 336 ft ^2 so the answer checks.

That's what I get, x = 8. (I didn't bother simplifying the negative solution.)

Small typo in your quadratic equation -- should be $4.5x^2 + 6x - 336 = 0$.

 

[hackedagainanda]

Thanks to everyone for your patience and help!