4. Find the dimensions of the largest isosceles triangle having a perimeter of 18 cm. Answer ( all
sides 6 cm)
i really need help with coming up with the equation of the function before solving its derivative and making it equal zero
so far i know 2x+y=18
To find the dimensions of the largest isosceles triangle with a perimeter of 18 cm, we can start by labeling the lengths of the equal sides of the triangle as x and the length of the base as y.
Since an isosceles triangle has two equal sides, the total length of the two equal sides would be 2x. Therefore, the equation for the perimeter of the triangle can be written as:
2x + y = 18
Now, we want to find the dimensions (x and y) that maximize the area of the triangle. The area of an isosceles triangle can be calculated using the following formula:
Area = (1/4) * √(4x^2 - y^2) * y
To maximize the area, we need to find the critical points of the area function by taking the derivative of the area with respect to y and setting it equal to zero.
Let's calculate the derivative of the area function:
d/dy [(1/4) * √(4x^2 - y^2) * y] = (1/4) * [2y * √(4x^2 - y^2) - (y^2 / √(4x^2 - y^2)]) = 0
Simplifying the equation above, we get:
2y * √(4x^2 - y^2) - (y^2 / √(4x^2 - y^2)) = 0
Multiplying both sides of the equation by √(4x^2 - y^2), we have:
2y * (4x^2 - y^2) - y^2 = 0
Expanding the equation further, we get:
8x^2y - 2y^3 - y^2 = 0
Rearranging the terms, we have:
8x^2y = 2y^3 + y^2
Simplifying the equation, we get:
4x^2 = y^2 + (1/4)y^2
Combining like terms, we have:
4x^2 = (5/4)y^2
We can now substitute the value of y in terms of x from the perimeter equation.
From the perimeter equation, we have:
2x + y = 18
Rearranging the equation, we get:
y = 18 - 2x
Substituting the value of y in terms of x into the area equation, we have:
4x^2 = [(5/4)*(18 - 2x)]^2
Now, we can solve this equation for x.
Expanding the equation further, we get:
4x^2 = (5/4)^2 * (18 - 2x)^2
Simplifying the equation, we get:
4x^2 = (25/16) * (18 - 2x)^2
Multiplying both sides of the equation by 16 to clear the fraction, we have:
64x^2 = 25 * (18 - 2x)^2
Expanding the equation, we get:
64x^2 = 25 * (324 - 72x + 4x^2)
64x^2 = 8100 - 1800x + 100x^2
Rearranging the terms, we have:
100x^2 - 1800x + (8100 - 64x^2) = 0
Finally, solving this quadratic equation for x will give you the values for the length of the equal sides of the triangle.
Please note that solving this equation may require further steps or approximations using quadratic formula or factoring.
To find the dimensions of the largest isosceles triangle, you can start by setting up an equation based on the given information. Let's assume the length of the two equal sides of the triangle is represented by 'x', and the length of the base (which is not equal) is represented by 'y'.
Given that the perimeter of the triangle is 18 cm, you can write the equation:
2x + y = 18
This equation represents the sum of the lengths of all three sides of the triangle.
To find the maximum value of the isosceles triangle, we need to find the values of 'x' and 'y' that satisfy this equation while maximizing the area of the triangle. The area of an isosceles triangle can be calculated using the formula:
Area = (1/4) * √(4x^2 - y^2) * y
To maximize the area, you need to take the derivative of the area equation with respect to 'y', set it equal to zero, and solve for 'y'. However, before proceeding with finding the derivative, let's simplify the equation:
2x + y = 18
=> y = 18 - 2x
Substituting this simplified equation for 'y' in the area formula gives:
Area = (1/4) * √(4x^2 - (18 - 2x)^2) * (18 - 2x)
Now, differentiate the area function with respect to 'x':
dA/dx = (1/4) * [(4x^2 - (18 - 2x)^2)^(-1/2)] * (8x - 36 + 4x)
= (1/4) * [(4x^2 - (324 - 72x + 4x^2))^(-1/2)] * (12x - 36)
Set dA/dx equal to zero to find the critical points:
(1/4) * [(4x^2 - (324 - 72x + 4x^2))^(-1/2)] * (12x - 36) = 0
Simplify and solve for 'x':
(12x - 36) = 0
12x = 36
x = 36/12
x = 3
Now that we have found the value of 'x', we can substitute it back into the simplified equation for 'y':
y = 18 - 2x
y = 18 - 2(3)
y = 18 - 6
y = 12
Thus, the dimensions of the largest isosceles triangle with a perimeter of 18 cm are: x = 3 cm and y = 12 cm, which means all sides of the triangle are 6 cm long.
For an isosceles triangles with equal legs x, and base y, the height is
√(x²-(y/2)²), where the base length is given by
y=18-2x .....(1)
The area if such a triangle is:
A(x)
=(1/2)base*height
=(y/2)*√(x²-(y/2)²)...(2)
Substitute the value of y from equation (1) to reduce equation (2) to a function of x only.
Differentiate A(x) with respect to x and equate A'(x) to zero to determine the value of x to maximize the area.
Check that A"(x) is negative, i.e. A is a maximum.
I get x=6 (equilateral triangle).