A farmer has 30yd of fencing and wants to enclose an area beside his barn. What are the dimensions of the region of maximal area, and what is that maximal area if the region is an is isosceles triangle with its base along side of the barn. Given -> legs of length 15, x = 1/2 base
To find the dimensions and maximal area of the region, we need to use the formula for the area of an isosceles triangle and the given information.
Let's start by visualizing the problem. The farmer wants to enclose an isosceles triangle with one side being the base along the side of the barn. This means two sides of equal length (legs) and one different side (base).
We are given that the legs have a length of 15 yards each, and the formula for the area of an isosceles triangle is:
Area = 1/2 * base * height
In this case, the base of the triangle is represented by "x," and the height is the perpendicular distance from the base to the top vertex of the triangle.
Given that x = 1/2 * base, we can substitute this value into the area formula:
Area = 1/2 * x * height
Now, let's calculate the height in terms of "x" using the Pythagorean theorem:
Since the base divides the isosceles triangle into two congruent right triangles, we can find the height using one of them.
Using the Pythagorean theorem, we have:
(leg)^2 + (height)^2 = (hypotenuse)^2
Substituting the values, we get:
15^2 + height^2 = x^2
225 + height^2 = x^2
Rearranging the equation, we get:
height^2 = x^2 - 225
height = √(x^2 - 225)
Now, we substitute this height value into the area formula:
Area = 1/2 * x * √(x^2 - 225)
To find the dimensions of the region of maximal area, we need to find the value of "x" that maximizes the area. We can do this by taking the derivative of the area formula with respect to "x" and setting it equal to zero:
(dA/dx) = 0
To simplify the formula before taking the derivative, you can multiply both sides of the equation by 2 to eliminate the fraction:
2 * Area = x * √(x^2 - 225)
Now, we can differentiate both sides of the equation with respect to "x":
(dA/dx) = √(x^2 - 225) + x * (1/2) * (2x) / √(x^2 - 225)
Simplifying the derivative:
0 = √(x^2 - 225) + x^2 / √(x^2 - 225)
Now, we can solve this equation for "x" by isolating the radical term:
x^2 / √(x^2 - 225) = -√(x^2 - 225)
Squaring both sides of the equation to eliminate the radical:
x^2 = √(x^2 - 225)^2
Expanding and simplifying:
x^4 - 450x^2 + 50625 = x^2 - 225
Rearranging the equation:
x^4 - 451x^2 + 50850 = 0
At this point, you can solve this equation using numerical methods or factoring techniques, or you can graph the equation and find the x-values where the curve intersects the x-axis.
let the base be 2x ,(if x = 1/2 the base)
and let each of the equal sides be y
so we know 2x + 2y = 30
x+y = 15, or y = 15-x
let the height of the triangle be h
then h^2 + x2 = y^2
h^2 + x^2 = (15-x)^2
h^2 + x^2 = 225 - 30x + x^2
h = √(225-30x)
area = (1/2)base x height
= (1/2)(2x)(√(225-30x) )
= x (225-30x)^(1/2)
d(area) = x(1/2)(225-30x)^(-1/2) (-30) + (225-30x)^(1/2)
=0 for a max of area
(225-30x)^(1/2) = 15x/(225-30x)^(1/2)
cross-multiply
15x = 225-30x
45x = 225
x = 5
then h = √(225-150) = √75
and since x+y = 15
y = 10
so each side of the triangle must be 10 yds
and the largest area is (1/2)(10)(√75)
= 25√3
Makes sense that the triangle of largest area would be an equilateral triangle, just like the largest quad would be a square, etc.