Trig/Pre-Calc- Finding Dimensions

[orlilovescamels]

My Trig teacher gave us three problems she got from a Pre-Calc course at some university. We are not allowed to graph equations with our calculators to get the answer. She also says that at some point we need a quadratic equation to be shown. Needless to say, I'm really lost.
Here's the first one:

"The following figure shows a square inscribed within a unit square. For which value of x is the area of the inner square a minimum? What is the minimum area?
Hint: Denote the lengths of the two segments that make up the base of the unit square by t and 1-t. Now use the Pythagorean theorem and congruent triangles to express x in terms of t."

So what I have written down so far is:
(Using Pythagorean theorem) x= the sq. root of the quantity (t^2 + (1-t)^2)
and
(Using Area of a Triangle):
x^2 = 1- 4(.5(t(1-t)))
= 1-2(t(1-t))
=1-2t+2t^2
=2t^2-2t+1
And then I plug that into the Quadratic Formula. BUT, the number I get under the radical is -4. Imaginary sq. roots are not good.

Can anyone tell me what went wrong, or at least point me in a better direction

 

[stapel]

....And then I plug that into the Quadratic Formula.

What, exactly, did you plug into the Quadratic Formula? As written, you could only have solved for x in terms of t, or for t in terms of x. But you could not have obtained a numerical answer. So some steps must be missing.

Please reply with that missing information. Thank you!

Eliz.

P.S. As drawn, you can apply the Pythagorean Theorem to the triangles in the corners. This will allow you to express the area of the inner square in terms of t; that is, the area expression A can be written as the function A(t), where A(t) is a quadratic. Then you can minimize the area A by finding the vertex of the quadratic function A(t).

 

[orlilovescamels]

....And then I plug that into the Quadratic Formula.

What, exactly, did you plug into the Quadratic Formula? As written, you could only have solved for x in terms of t, or for t in terms of x. But you could not have obtained a numerical answer. So some steps must be missing.

Please reply with that missing information. Thank you!

Eliz.

P.S. As drawn, you can apply the Pythagorean Theorem to the triangles in the corners. This will allow you to express the area of the inner square in terms of t; that is, the area expression A can be written as the function A(t), where A(t) is a quadratic. Then you can minimize the area A by finding the vertex of the quadratic function A(t).

I put 2t^2-2t+1, plugging in 2 for a, -2 for b, and 1 for c.
The answer was supposed to come out to be the area of the inner square.
The square root of that coming out to x.

And I know the pythag. theorem can get me x; x= sq. rt. of (t^2 + (1-t)^2)
But I have no clue how to solve for a variable using variables.
And, I don't know how to find the vertex without a calculator unless you're saying to use y-k=a(x-h)^2, where (h,k) is vertex.

I hope that's what you needed to know.

 

[pka]

Each of the triangles has an area of $\displaystyle \frac{{t\left( {1 - t} \right)}}{2}$.
There are four of them. So the internal square has an area of:
$\displaystyle 1 - 2t\left( {1 - t} \right)$.
If you maximize $\displaystyle 2t\left( {1 - t} \right)$ then you minimize the area of the smaller square.

 

[orlilovescamels]

Each of the triangles has an area of $\displaystyle \frac{{t\left( {1 - t} \right)}}{2}$.
There are four of them. So the internal square has an area of:
$\displaystyle 1 - 2t\left( {1 - t} \right)$.
If you maximize $\displaystyle 2t\left( {1 - t} \right)$ then you minimize the area of the smaller square.

So tell me how to maximize that without using a calculator.

 

[soroban]

Hello, orlilovescamels!

The following figure shows a square inscribed within a unit square.
For which value of x is the area of the inner square a minimum?
What is the minimum area?
Hint: Denote the lengths of the two segments that make up the base of the unit square by t and 1-t.
Now use the Pythagorean theorem and congruent triangles to express x in terms of t.

We want to minimize the area of the inner square: $\displaystyle \:A \:=\:x^2$

From the right triangles, we have: $\displaystyle \:x^2\:=\:t^2\,+\,(1-t)^2 \:=\:2t^2\,-\,2t\,+\,1$

. . And that is the area function: $\displaystyle \:A \;=\;2t^2\,-\,2t\,+\,1$


How do we minimize the value of $\displaystyle A$ ?

Note that the equation is a parabola, opening upward.
Its minimum occurs at its vertex: $\displaystyle \:v \:=\:\frac{-b}{2a}$

We have: $\displaystyle \,a\,=\,2,\;b\,=\,-2$
. . Hence, the vertex is at: $\displaystyle \:t \:=\:\frac{-(-2)}{2(2)} \:=\:\frac{1}{2}$

To get the inner square with minimum area, use the midpoints of the sides.


From the right triangles, we have: $\displaystyle \:x^2\:=\:\left(\frac{1}{2}\right)^2 \,+\,\left(\frac{1}{2}\right)^2 \:=\:\frac{1}{4}\,+\,\frac{1}{4} \:=\:\frac{1}{2}$

Therefore, the area of the inner square is: $\displaystyle \:A \:=\: x^2\:=\:\frac{1}{2}$

 

[pka]

That is a parabola. It has max/min at the vertex.

 

[orlilovescamels]

Hello, orlilovescamels!

The following figure shows a square inscribed within a unit square.
For which value of x is the area of the inner square a minimum?
What is the minimum area?
Hint: Denote the lengths of the two segments that make up the base of the unit square by t and 1-t.
Now use the Pythagorean theorem and congruent triangles to express x in terms of t.

We want to minimize the area of the inner square: $\displaystyle \:A \:=\:x^2$

From the right triangles, we have: $\displaystyle \:x^2\:=\:t^2\,+\,(1-t)^2 \:=\:2t^2\,-\,2t\,+\,1$

. . And that is the area function: $\displaystyle \:A \;=\;2t^2\,-\,2t\,+\,1$


How do we minimize the value of $\displaystyle A$ ?

Note that the equation is a parabola, opening upward.
Its minimum occurs at its vertex: $\displaystyle \:v \:=\:\frac{-b}{2a}$

We have: $\displaystyle \,a\,=\,2,\;b\,=\,-2$
. . Hence, the vertex is at: $\displaystyle \:t \:=\:\frac{-(-2)}{2(2)} \:=\:\frac{1}{2}$

To get the inner square with minimum area, use the midpoints of the sides.


From the right triangles, we have: $\displaystyle \:x^2\:=\:\left(\frac{1}{2}\right)^2 \,+\,\left(\frac{1}{2}\right)^2 \:=\:\frac{1}{4}\,+\,\frac{1}{4} \:=\:\frac{1}{2}$

Therefore, the area of the inner square is: $\displaystyle \:A \:=\: x^2\:=\:\frac{1}{2}$

Oooh. I'm so embarassed that I forgot the vertex equation.
Thank you so much!

 

[Denis]

Hmmm....A = x^2 = 2t^2 - 2t + 1

Set to zero:
2t^2 - 2t + 1 = 0

take 1st derivative:
4t - 2 = 0

t = 1/2

Too simple?

 

[stapel]

take 1st derivative:....

She's in pre-calculus, so she probably doesn't know about derivatives, yet.

In any case, she didn't have to bother with following the step-by-step instructions (finding the area A in terms of t by using the right triangles, because of t's relation to x and the area x², and then finding the minimum by finding the vertex), because Soroban did it for her. :roll:

Eliz.

 

[stapel]

I put 2t^2-2t+1, plugging in 2 for a, -2 for b, and 1 for c....

The Quadratic Formula requires that you have "ax² + bx + c = 0". What did you have that equalled zero? What were your steps that justified using the Quadratic Formula? What, exactly, was your reasoning? :?:

Since there was no basis for assigning the area the value of zero, it might have been useful to try following the step-by-step instructions that were provided:

P.S. As drawn, you can apply the Pythagorean Theorem to the triangles in the corners. This will allow you to express the area of the inner square in terms of t; that is, the area expression A can be written as the function A(t), where A(t) is a quadratic. Then you can minimize the area A by finding the vertex of the quadratic function A(t).

Soroban won't always be available to follow the instructions and provide a complete worked solution for you, so it might be helpful, in the future, to give it a try yourself! :wink:

Eliz.