find the greatest area..please help
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Assume that the Sides are making angle Θ with the wall.
Now calculate the area in terms of Θ and maximize.
Please share your work with us .
If you are stuck at the beginning tell us and we'll start with the definitions.
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sorry, I read the rules but I had already posted this. I added my work so far to the other problem I had posted, and will too with this one but I am stuck at the beginning on this one.....I don't know what that little symbol means, we haven't gone over that yet..? is there another way?
thank you for your time
thank you for your time
The name of symbol θ is "theta" (pronounced THAY-tuh). Theta is the eighth letter of the Greek alphabet. Mathematicians often use letters of the Greek alphabet to represent angle measures. This is something that you will learn, if you study trigonometry.
Yes.
Many different isosceles triangles may be formed with two 16-foot sections of fence and a wall. It's the height of the triangle and the length of the base that vary. (The greater the height, the narrower the base -- the shorter the height, the wider the base. Can you envision this?)
Let h = the height of the triangle
Now calculate the half-length of the base, in terms of h.
What do you get?
Cheers :cool:
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I cant find a without h, and visa versa? I divided the triangle straight down the middle, calling that line (height) h. I was then going to solve for a using the Pythagorean theorem. but I still have two unknowns. I used b for hypotenuse and a for base (wall side) and came up with h^2 +a^2= b^2
then a^2=16^2-h^2
sq.rt. of ^2= sq.rt. of 16^2-h^2
a= sq. rt. of 256-h^2
do I substitute this into the area= 1/2ah equation?
then a^2=16^2-h^2
sq.rt. of ^2= sq.rt. of 16^2-h^2
a= sq. rt. of 256-h^2
do I substitute this into the area= 1/2ah equation?
Dividing the isosceles triangle in half vertically results in two, identical right triangles.
You're calling a the base of each of those right triangles, yes? (I would like to confirm that you're not using a to represent the length of the wall.)
This looks good, as long as you're working with only half the garden triangle, here.
Let's summarize what we have so far.
1) height of the garden triangle is h
2) one-half of the base of the garden triangle is sqrt(256 - h^2)
3) let A = the area of the garden triangle
A = ?
Once you have the area formula, do you plan on maximizing A by using the same graphical method as in your other exercise?
You are working with more than one triangle. I would like to confirm your meaning, when you say that a is the base.
Is your a the base of the garden triangle, or is it the base of the right triangles?
This is not in terms of h; it's in terms of a.
It's also not what I asked you to do. (You were correct the first time, if your a is half the base of the garden triangle.)
I can understand why!
The Pythagorean Theorem applies only to right triangles.
You may not use this theorem on an isosceles triangle, or an equilateral triangle, or any other kind of triangle that does not contain a right angle.
So, after dividing the garden triangle in half, you have two right triangles, yes?
We can work with one of them because they are both identical.
The height of this right triangle is h
The base of this right triangle is a
The hypotenuse of this right triangle is 16
Your earlier result still applies, by the Pythagorean Theorem.
a = sqrt(256 - h^2)
In other words, 1/2 the base of the garden triangle is sqrt(256-h^2), and the height of the garden triangle is h.
What is the area formula for the garden triangle?
A =
That's one way to think about it. (The factor of 2 on top cancels with the factor of 2 on the bottom, by the way.)
Here's another way to think about it.
The area of a triangle is 1/2 the base times the height.
You chose a symbol for the height; h
You calculated half the base in terms of h; sqrt(256-h^2)
So just multiply them together!
A = sqrt(256 - h^2)*h
Same result as yours, once you cancel your common factors.
Are you planning on finding the maximized A the same way as before (by graphing)?
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You are now using undefined symbols.
I'm going to force you to define your symbols x and y, before I respond again.
(Mathematics requires very precise language; we must be thorough, when communicating about math.)
Okay. x is the area of the garden, and y is the height of the isosceles triangle.
In your previous post, you stated the value of x as 11.32 and the value of y as 128.
Would you like to revise anything, at this point?