Question

An isosceles triangle has sides of length x,x and y. In addition, assume the triangle has perimeter 12.
a) Find the rule for a function that computes the area of the triangle as a function of x. Describe the largest possible domain of this function.

Answer 1

The first thing to do is to draw a diagram


Perimeter is 12, so you can say that 2x+y=12

Now, let's move to the area.

Area of a triangle is 1/2 bh
Let b=y from the diagram. Now, we must find h.

If you draw an imaginary line for h that acts like a perpendicular bisector for base y, you can relate the base, y, and x using the Pythagorean thm. You can conclude that
\(h=\sqrt{x^2 - (\frac{y}{2})^2}\)



Hence, Area of the full isosceles triangle is

\(A=(\frac{1}{2})(y)(\sqrt{x^2 - (\frac{y}{2})^2})\)
The issue we have is that there are 3 variables which is unfavorable for the Cartesian plane. Hence, manipulation is required.

Recall the equation for perimeter
2x+y=12
You can say that y=12-2x
Substitute the expression for y into the area.
\(A=(\frac{1}{2})(y)(\sqrt{x^2 - (\frac{y}{2})^2}\)
\(y=12-2x\)
Hence,
\(A=(\frac{1}{2})(12-2x)(\sqrt{x^2 - (\frac{12-2x}{2})^2}\)

That was fun

Now for the maximum, you can either use a graphing calculator or take the derivative \(\frac{dA}{dx}\) and find critical points. Then evaluate if they are a maximum by using the first or second derivative test.

Answer 2

Oh wait it said domain.

The only limiting thing here is the square root. Set the inside to 0 and solve for x. That is the lowest x value possible.

Answer 3

whats a derivative

Answer 4

we are not allowed to use graphing calculators either

Answer 5

Derivative is a way to measure the instantaneous rate of change of a function, where the point of interest is continuous along f(x) and that \(\lim_{h\rightarrow{0}}\frac{f(h+a)-f(a)}{h}=f'(x)\)

this might explain better:https://www.youtube.com/watch?v=lowavG2SXsQ


But it's not required for this question, I misunderstood.

Answer 6

Wait, so by 'largest possible domain' of this function is it asking for the domain interval or...?

Answer 7

brb in a min

Answer 8

the answer is domain 3 ≤
x ≤ 6

Answer 9

so domain interval ithink

Answer 10

Yes that is correct.

Sorry, so since this is a triangle, negative areas or not possible. Hence the function we found for A(x) is only bounded by where A(x) is above the x axis. You can solve for the zeroes of the function and determine by plugging in if the intervals between the zeroes are positive or negative.

Answer 11

thx

Answer 12

I got the domain to be from -9 to 6 where it is non-zero. But the answer is 3 to 6. What did I do wrong?

Answer 13

I set -2x+12=0 and x+9 = 0

Answer 14

Where did you get x+9 = 0?

Answer 15

so I have (1/2)(y)(sqrt(x^2-(y/2)^2) right?
I plugged -2x+12 for y.
so Now I have (-x+6)(sqrt(12x+36))
sqrt(12x+36) is the same as sqrt((4)(x+9))
so I have (-x+6) times 2 times sqrt(x+9)

Answer 16

wait, I think I did math wrong

Answer 17

so I got sqrt(12x-36)

Answer 18

nvm I got it, thanks

Answer 19

Ah okay, glad you got it (:

Answer 20

I'm a little confused on the second part:
Assume that the maximum value of the function a(x) in (a) occurs when x=4.Find the maximum value of z=a(x) and z=2a(3x+3) +1.

Answer 21

I got the first part

Answer 22

I don't get the second part though

Answer 23

my brain hurts

Answer 24

lmao I did all that work for nothing my dude....

Basically you need to take the maximum you got for \(z=a(x)\) and plug that into \(2(n)+1\)

Here's why. Let's say you have a function \(f(x)\) which has an extremum at \(x=n\). If you now have \(f(g(x))\) the extremum will still be at \(x=n\).

By this, the maximum you got for \(a(x)\) at \(x=n\) will just need to be plugged in to \(2n+1\).

Answer 25

my brain also hurts

Answer 26

this shouldn't be a calculus problem

Answer 27

So the maximum for \(a(x)\) is y=\(4\sqrt{3}\)

Just find
\(2(4\sqrt{3})+1\)

Answer 28

yes

Answer 29

No it isn't, it just happens when your common sense goes to sleep :/

Answer 30

so the answer for finding the max value for z=2a(3x+3) +1. is just 2(max for a(x)) +1?

Answer 31

wow!

Answer 32

Yes. The maximum for \(f(g(x))\) is the same as the maximum for \(f(x)\) because putting a composite function inside doesn't move the function up or down.

Answer 33

thanks

Answer 34

this is part 3...

Answer 35

I drew the original graph and applied the transformations. But how does the rule of the function transform in 2a(3x+3)+1?

Answer 36

nvm, I think I got it

Answer 37

nope, I don't got it

Answer 38

I did this..

so I have (-2x+12)(sqrt(3x+3)) right?
then I got 2((-2(3x+3)+12)(sqrt(3(3x+3)+1)))+1

Answer 39

I got it, I just did math wrong