What values of x will result in combined areas of squares exceeding 5 cm^2?

[mathdad]

A piece of wire is cut into two pieces. Denote the lengths of the two pieces by x and 12- x. Both pieces are then bent into squares. For which values of x will the combined areas of the squares exceed 5 cm^2?

This question is from the quadratic inequality section of the textbook. I assume an inequality set up is needed.
Can someone set up the needed inequality and provide the steps to solve this inequality application?

 

[MarkFL]

What is the area $A$ of a square whose perimeter is $P$?

 

[mathdad]

Perimeter of square is P = 4(side).

Area of square = (side)^2.

What's the next step?

 

[MarkFL]

Perimeter of square is P = 4(side).

Area of square = (side)^2.

What's the next step?

Let's use x for the length of the sides of the square, so we have:

\(\displaystyle P=4x\)

\(\displaystyle A=x^2\)

Now, we want to replace x in the formula for the area with an expression containing P...:D

 

[mathdad]

P = 4x

P/4 = x

A = x^2

A = (P/4)^2

A = P^2/16

What is next?

 

[MarkFL]

P = 4x

P/4 = x

A = x^2

A = (P/4)^2

A = P^2/16

What is next?

Okay, you have a piece of wire whose length is let's say W...you bend it into a square...what is the perimeter of the square? Then using your formula for the area of a square as a function of its perimeter, what is the area of the square as a function of W?

 

[mathdad]

A = p^2/16

A(w) = w^2/16

 

[MarkFL]

A = p^2/16

a(w) = w^2/16

Yes, now we need to answer the question:

For which values of x will the combined areas of the squares exceed 5 cm^2?

We have two pieces of wire whose lengths are given...what is the combined area of the squares resulting from bending the two pieces into squares? What mathematical statement can we use to determine the answer to the above question we wish to answer?

 

[mathdad]

Let x = side of first square

A = x^2

Let (12 - x) = side of second square

A = (12 - x)^2

Combining the two areas we get

x^2 + (12 - x)(12 - x)

x^2 + 144 - 24x +x^2

2x^2 - 24x + 144

We must use the math statement less than or equal to.

2x^2 - 24x + 144 ≤ 5 cm^2

Correct?

 

[MarkFL]

Let x = side of first square

A = x^2

Let (12 - x) = side of second square

A = (12 - x)^2

Combining the two areas we get

x^2 + (12 - x)(12 - x)

x^2 + 144 - 24x +x^2

2x^2 - 24x + 144

We must use the math statement less than or equal to.

2x^2 - 24x + 144 ≤ 5 cm^2

Correct?

No, what we want is:

\(\displaystyle A\left(W_1\right)+A\left(W_2\right)>5\)

\(\displaystyle A(x)+A(12-x)>5\)

Now, the arguments for the area function represent wire lengths, and so using our function we found for the area of a square resulting from bending a wire of a given length into a square, we have:

\(\displaystyle \frac{x^2}{16}+\frac{(12-x)^2}{16}>5\)

Now, solve for x and keep in mind that x and 12 - x represent lengths, and so cannot be negative. :D

 

[mathdad]

I can solve this quadratic inequality and will do so when I get home. I want to thank you for guiding me one step at a time through each posted question. I am learning to reason my way to the answer and/or formula.

 

[mathdad]

(x^2)/16 + (12 - x)^2/16 > 5

(x^2)/16 + (12 - x)^2/16 - 5 > 0

After much paper work, I got the following:

(2x^2 - 24x + 64)/16 > 0

I set the numerator to 0. I found x = 4 and x = 8 to be the values of x.

Questions:

Did I find the correct values of x to answer this question?

Do I also need to find the solution of the quadratic inequality?

 

[MarkFL]

(x^2)/16 + (12 - x)^2/16 > 5

(x^2)/16 + (12 - x)^2/16 - 5 > 0

After much paper work, I got the following:

(2x^2 - 24x + 64)/16 > 0

At this point you can multiply through by 8 to get:

\(\displaystyle x^2-12x+32>0\)

Then factor:

\(\displaystyle (x-4)(x-8)>0\)

Now solve this inequality, subject to:

\(\displaystyle 0<x<12\)

You understand why we need this constraint, right?

 

[mathdad]

Explain why we need the constraint.

Are you saying to solve (x - 4) (x - 8) > 0?

 

[mathdad]

x = 4

x = 8(x - 4)(x - 8) > 0<----------(4)---------------(8)---------->

I can see that we must reject x = 4 and x = 8.

For (-infinity, 4), let x = 0. Here we have a true statement.

For (4, 8), let x = 5. Here we have a false statement.

For (8, infinity), let x = 9. Here we have a true statement.

Solution: (-infinity, 4) U (8, infinity).

Correct?

How does this solution answer the question about the two wires?

 

[MarkFL]

Explain why we need the constraint.

x and x - 12 represent lengths (assumed to be in cm)...and so must be positive. Lengths cannot be negative, and if we are to actually have two pieces of wire, then these lengths must both be positive.

Are you saying to solve (x - 4) (x - 8) > 0?

x = 4

x = 8(x - 4)(x - 8) > 0<----------(4)---------------(8)---------->

I can see that we must reject x = 4 and x = 8.

For (-infinity, 4), let x = 0. Here we have a true statement.

For (4, 8), let x = 5. Here we have a false statement.

For (8, infinity), let x = 9. Here we have a true statement.

Solution: (-infinity, 4) U (8, infinity).

Correct?

Now you need to apply the constraint...so that we have:

\(\displaystyle (0,4)\,\cup\,(8,12)\)

Any value of x in the above intervals will result in the combined area of the squares formed by bending the two pieces of wire into square being greater than 5 cm². :D

 

[mathdad]

Thank you for completing the problem. I do know that length cannot be negative. Tell me, where did 0 and 12 come from? I know you mean 0 < x < 12 but is there a connection between 0 < x < 12 and the two lengths given in the word problem?

 

[MarkFL]

Thank you for completing the problem. I do know that length cannot be negative. Tell me, where did 0 and 12 come from? I know you mean 0 < x < 12 but is there a connection between 0 < x < 12 and the two lengths given in the word problem?

We require both:

\(\displaystyle 0<x\)

\(\displaystyle 0<12-x\)

Combined, this gives us:

\(\displaystyle 0<x<12\)

Perhaps a more intuitive way to look at it is to imagine we have laid the straight wire before us, horizontally. Beginning at the left end, we then move x units to the right to make our cut, cutting the wire into two pieces. Because we need 2 pieces of wire, we have to move some positive distance (0 < x) and since the wire is 12 cm in length, we have to move some distance less than 12 cm (x < 12). In other words, the distance x that we can move to make our cut is:

\(\displaystyle 0<x<12\)

 

[mathdad]

Ok. This is it for today. I may post two more quadratic inequality questions tomorrow before moving on to the next section.