Max/Min Triangle Area in Calc BC: Find Dimensions of Isosceles Triangle

[Albert Einstein]

This is in my Calc BC class, a homework question from the maximums and minimums section. Can someone solve it please? I know the answers but my work doesn't look too good. Thanks a lot.


Find the dimensions of the isosceles triangle of least area that can be cirumscribed about a circle of radius R.

the answers are:
height= 3R
base=2R(3)^(1/2)
sides=2R(3)^(1/2)

 

[HallsofIvy]

Was there a reason for posting this under differential equations? It would be better in the Homework section. Oh- and while your work may not "look too good", it would help if you showed what work you did.

 

[M98Ranger]

Sure, I can help you solve this problem. First, let's draw a diagram to visualize the problem.

We have a circle with radius R, and we want to inscribe an isosceles triangle around it. Let's label the center of the circle as point O, and the points where the triangle touches the circle as A, B, and C. Since the triangle is isosceles, we know that the sides AB and AC are equal, and the angle at the top of the triangle (angle BAC) is also equal to the angles at the base (angle ABC and angle ACB).

Next, let's label the height of the triangle as h, and the base as b. We can use the Pythagorean theorem to find the length of the sides AB and AC. Since the radius of the circle is R, we know that the distance from point O to either point A, B, or C is also R. Therefore, we can write the following equation:

(R)^2 + (h)^2 = (AB)^2

We also know that the distance from point A to point B is equal to the distance from point A to point C (since the triangle is isosceles). So we can write another equation:

(b/2)^2 + (h)^2 = (AC)^2

Now, we can use the formula for the area of a triangle (A = 1/2 * base * height) to find the area of this triangle. Since we want to minimize the area, we can set up a function for the area in terms of h and b, and then use differentiation to find the critical points.

A(h,b) = 1/2 * b * h

Next, we can substitute the values for AB and AC into our function:

A(h,b) = 1/2 * (2R(3)^(1/2)) * h

Now, we can substitute the value for h from our first equation into our function:

A(b) = 1/2 * (2R(3)^(1/2)) * sqrt((AB)^2 - (R)^2)

We can simplify this function to:

A(b) = R(3)^(1/2) * sqrt(b^2 - 4R^2)

Finally, we can use differentiation to find the critical points of this function:

A'(b) = R(3)^(

 

[M98Ranger]

Sure, I would be happy to help solve this problem for you. First, let's start by drawing a diagram to better understand the situation. We have a circle with a radius of R, and we want to inscribe an isosceles triangle around it with the least possible area.

Since we know that an isosceles triangle has two equal sides, we can label the base as x and the two equal sides as y. The height of the triangle can be found using the Pythagorean theorem, which states that the square of the hypotenuse (in this case, the radius of the circle) is equal to the sum of the squares of the other two sides.

So, we have:
(R)^2 = (x/2)^2 + y^2

We can simplify this to:
4R^2 = x^2 + 4y^2

Next, we need to find the area of the triangle. The formula for the area of a triangle is 1/2 * base * height. In this case, the base is x and the height is y, so the area can be written as:
A = 1/2 * x * y

We can substitute the value of y from our previous equation into this formula, giving us:
A = 1/2 * x * √(4R^2 - x^2)

Now, we can use calculus to find the minimum value of this function. We will take the derivative of A with respect to x, set it equal to 0, and solve for x.

dA/dx = 1/2 * (√(4R^2 - x^2) - x * (4R^2 - x^2)^(-1/2) * (-2x))

Setting this equal to 0 and solving for x gives us:
x = √(4R^2 - x^2)

Squaring both sides and simplifying, we get:
x^2 = 4R^2 - x^2

Solving for x gives us:
x = 2R√3

Now, we can plug this value back into our equation for y to find the height of the triangle:
(R)^2 = (x/2)^2 + y^2
(R)^2 = (2R√3/2)^