Calculus Trig A triangle is inscribed inside a semi-circle with radius of 2 find maximum area

[Skh]

need help im just stuck I tried everything I'm just confused I know area is 4 a is 4 b is 86 and c is 90 and a formula could be y^2+x^2=4 but I dont know anything else

 

[Deleted member 4993]

need help im just stuck I tried everything I'm just confused I know area is 4 a is 4 b is 86 and c is 90 and a formula could be y^2+x^2=4 but I dont know anything else View attachment 13944

Please explain further:

You write: "I know area is 4 a is 4 b is 86 and c is 90" - what are 'a' & 'b' and 'c' as related to the diagram given. Which area are you referring to?

 

[Deleted member 4993]

Let A and B be the diametrically opposite point and let C be the other vertex.

Let the co-ordinate of C be (x₁,y₁)

The area of the triangle ABC = 1/2 * (AB) * y₁ = 2*y₁

Knowing that C is on a circle of radius 2 - what is the range y₁?

What is the maximum value of the area?

 

[MarkFL]

I'd be inclined to let the circle have radius $r$, and the vertex of the triangle be at the point:

[MATH](r\cos(t),r\sin(t))[/MATH] where $0\le t\le\pi$

And so the area $A$ of the triangle is:

[MATH]A(t)=\frac{1}{2}(2r)r\sin(t)=r^2\sin(t)[/MATH]
Can you proceed?

 

[Skh]

Please explain further:

You write: "I know area is 4 a is 4 b is 86 and c is 90" - what are 'a' & 'b' and 'c' as related to the diagram given. Which area are you referring to?

Area of the triangle being bh/2 so b would be 4 and h would be 2 from plugging it into the distance formula for the circle the area would be 4 from x^2+y^2 but if we are looking on the basis of the graph given the maximum point of the inscribed triangle would be (0,2)

 

[Skh]

I'd be inclined to let the circle have radius $r$, and the vertex of the triangle be at the point:

[MATH](r\cos(t),r\sin(t))[/MATH] where $0\le t\le\pi$

And so the area $A$ of the triangle is:

[MATH]A(t)=\frac{1}{2}(2r)r\sin(t)=r^2\sin(t)[/MATH]
Can you proceed?

What would t be? Base of the triangle?

 

[MarkFL]

What would t be? Base of the triangle?

$t$ is a parameter that allows the vertex on the circular arc to move counterclockwise from the point $(r,0)$ to the point $(-r,0)$. Here is a live graph to illustrate:

Desmos | Graphing Calculator

www.desmos.com

On the left, under the equations, you'll see a slider that you can move from $0$ to $\pi$. Move that slider to see all the different possible triangles.

 

[Deleted member 4993]

Area of the triangle being bh/2 so b would be 4 and h would be 2 from plugging it into the distance formula for the circle the area would be 4 from x^2+y^2 but if we are looking on the basis of the graph given the maximum point of the inscribed triangle would be (0,2)

Look at response #3.

The area of the triangle ABC = 1/2 * (AB) * y₁ = 2*y₁

Since (x1,y1) is on the circle - the maximum value of y₁ = r = 2

The maximum area of the triangle ABC = 1/2 * (AB) * [y₁]_max = 2*y₁ = 2 * 2 = 4

 

[Steven G]

Area of the triangle being bh/2 so b would be 4 and h would be 2 from plugging it into the distance formula for the circle the area would be 4 from x^2+y^2 but if we are looking on the basis of the graph given the maximum point of the inscribed triangle would be (0,2)

Why would h be 2? The arbitrary point (x₁,y₁) does NOT have be at the very top of the circle which will make h=2. y₁ can be lower. In this case h=y₁. Do you see that??

 

[Skh]

What would t be? Base of the triangle?

Cubed root (x^4) > 1/4(x^4) ^(-2/3) *4x^3 > 1/(3* cubed root(x^4)) * 4x^3 > (4/3 x)^(4/3)* x^2 > 4/3 x^(-4/3 + 2)> (4/3 x)