Math Help - Isosceles triangle - maximum area

1. Isosceles triangle - maximum area

An isosceles triangle has two equal sides of length 10 cm. Theta is the angle between two equal sides.
a) Express area of a triangle as a function of theta
b) If theta is increasing at a rate of 10 degrees/minute, how fast is area changing at the instant theta=pi/3?
c) at what value of theta will the triangle have the maximum area?

Thanks!!!

2. My answer is going to suck because I don't know how to use LaTex. I will try my best.

The first thing to do is to visualize the triangle in question. So I have an iso-triangle, with sides 10, and a base. Drawing a line from the apex to the base, we obviously see we have a right-triangle, with sides 10 (from the iso triangle), adjacent to $\theta$ would be side $X$, and opposite $\theta$ would be $\sqrt{(100-X^2)}$ (by the Pythagoreon theorom).

Now, the are of a triangle is $\frac{1}{2}(base)(height)$. This is the equation that is going to help us solve all the problems.

A) We want the triangles area expressed in terms of "theta" only. However our initial equation, using $\frac{1}{2}(base)(height)$ is:

$\frac{1}{2}(2X)(\sqrt{100-X^2})$

where

$2X$

is the base of the ENTIRE triangle (we defined $X$ is the base of one HALF of the triangle), and

$\sqrt{100-X^(2)}$

is the height of the triangle, which we defined by the Pythagorean theorem from the dimensions we had. However we have a problem. The question asks to define the equation in terms of $\theta$ and not $X$. So we have to come up with a relationship between $\theta$ and $X$.

We notice,

$\cos(\theta) = \frac{X}{10}$

by the dimentions of our triangle. Solving for

$X$,

we get $10\cos(\theta) = X$.

We can then plug in

$10\cos(\theta)$

everwhere we see $X$ in our original area formula to get:

$A(\theta) = \frac{1}{2}(2*10 \cos(\theta))(\sqrt{100-(10\cos(\theta))^2})$.

Now for the rest, I will leave that to you however:

B) We are trying to figure out the rate of change, so we're probably going to be taking some derivatives here to get $\frac{d(A(\theta))}{d(\theta)}$. We are also going to using implicit differentiation, and hopefully you can immediately spot where.

C) More derivatives, but here we are going to be looking for Maximum (and minimum if you want to test yourself) values.

3. Originally Posted by sun1

An isosceles triangle has two equal sides of length 10 cm. Theta is the angle between two equal sides.
a) Express area of a triangle as a function of theta
b) If theta is increasing at a rate of 10 degrees/minute, how fast is area changing at the instant theta=pi/3?
c) at what value of theta will the triangle have the maximum area?

Thanks!!!
See picture

$AB = BC = 10{\text{ cm}}{\text{, }}AH = HC = \frac{{AC}}{2}.$

According to a Law of cosines we have

$A{C^2} = A{B^2} + B{C^2} - 2AB \cdot AC\cos \theta = 100 + 100 - 2 \cdot 10 \cdot 10\cos \theta =$

$= 200 - 200\cos \theta = 200\left( {1 - \cos \theta } \right).$

According to the Pythagorean theorem we have

$B{H^2} = A{B^2} - {\left( {\frac{{AC}}{2}} \right)^2} = 100 - \frac{{200\left( {1 - \cos \theta } \right)}}{4} =$

$= 100 - 50\left( {1 - \cos \theta } \right) = 50\left( {1 + \cos \theta } \right).$

So, we have the following area of a triangle as a function of theta

${S_\vartriangle }\left( \theta \right) = \frac{1}{2}\sqrt {200\left( {1 - \cos \theta } \right)} \cdot \sqrt {50\left( {1 + \cos \theta } \right)} = 50\sqrt {1 - {{\cos }^2}\theta } = 50\sin \theta.$

As the sinus can vary from minus one to plus one, then, consequently, the maximum area of a triangle could be 50, i.e. ${S_\vartriangle }{\left( \theta \right)_{\max }} = 50$.

4. Originally Posted by sun1

An isosceles triangle has two equal sides of length 10 cm. Theta is the angle between two equal sides.
a) Express area of a triangle as a function of theta
b) If theta is increasing at a rate of 10 degrees/minute, how fast is area changing at the instant theta=pi/3?
c) at what value of theta will the triangle have the maximum area?

Thanks!!!
a) Drop a perpendicular from the vertex to the opposite side. That divides the isosceles triangle into two right triangles each having angle $\theta/2$ and hypotenuse 10. The "near side", the altitude of the isosceles triangle, is $10 cos(\theta)$ and the "opposite side", half the base of the isosceles triangle, is $10 sin(\theta)$ so the length of the base is $20 sin(\theta)$ and the area of the triangle is (1/2)(base)(height)= $(1/2)(10 cos(\theta))(20 sin(\theta))= 100 sin(\theta)cos(\theta)$.

b) Differentiate $A(\theta)= 100 sin(\theta)cos(\theta)$ with respect to $\theta$ and set $\theta= \pi/3$.

c) Set the derivative equal to 0 and solve for $\theta$.

5. Commenting on what DeMath did.
There is a nice theorem that says: The area of a triangle is $A_{\Delta}=\frac{a\cdot b\cdot \sin\theta}{2}$ where $a~\&~b$ are the lengths of two sides of the triangle and $\theta$ is the angle between them.

Thus we get $A_{\Delta}=50\sin(\theta)$. (note we do not need absolute value).

From that the rest of the question is done quite easily.