the hypotenuse of a right triangle is 12 cm. find one the legs if the area is maximum?
Intuitively, one should conclude that the sides would be equal, let's prove it
let one side by x, then the other is √(144 - x^2)
area = (1/2)x√(144-x^2) = (1/2)x(144-x^2)^(1/2)
d(area)/dx
= (1/2)[ x(1/2)(144-x^2)^(-1/2) (-2x) + (144-x^2)^(1/2) ]
= 0 for a max of area
-x^2/√(144-x^2) + √(144-x^2) = 0
x^2/√(144-x^2) = √(144-x^2)
x^2 = 144-x^2
2x^2 = 144
x^2 = 72
x = √72 = 6√2
one side is 6√2 , and the other side is
√(144 - 72) = √72 = 6√2
as anticipated.
check:
(√72)^2 + (√72)^2
= 144
tnx
To find the leg of a right triangle that results in the maximum area, we can use the Pythagorean theorem and calculus.
Let's assume that one of the legs of the right triangle is a, and the other leg is b. We know that the hypotenuse c is 12 cm.
According to the Pythagorean theorem, the relationship between the sides of a right triangle is:
a^2 + b^2 = c^2
Substituting the given values:
a^2 + b^2 = 12^2
a^2 + b^2 = 144
To find the maximum area, we need to maximize the function A = (1/2) * a * b. We can express b in terms of a using the equation of the hypotenuse:
a^2 + b^2 = 144
b^2 = 144 - a^2
b = sqrt(144 - a^2)
Now we can write the area function as:
A(a) = (1/2) * a * sqrt(144 - a^2)
To find the maximum area, we need to find the critical points of the area function. We can do this by taking the derivative of A(a) with respect to a and setting it equal to zero:
A'(a) = (1/2) * [sqrt(144 - a^2) - (a^2) / sqrt(144 - a^2)]
Setting the derivative equal to zero and solving for a:
0 = (1/2) * [sqrt(144 - a^2) - (a^2) / sqrt(144 - a^2)]
0 = sqrt(144 - a^2) - (a^2) / sqrt(144 - a^2)
Simplifying further:
(a^2) / sqrt(144 - a^2) = sqrt(144 - a^2)
a^2 = 144 - a^2
2a^2 = 144
a^2 = 72
a = sqrt(72)
a ≈ 8.49 cm
So, one of the legs of the right triangle is approximately 8.49 cm in length in order to maximize the area.
To find the leg of a right triangle that maximizes its area, we need to use calculus.
Let's denote one leg of the triangle as x cm. Then the other leg can be expressed as 12 - x cm, according to the Pythagorean theorem.
The area A of a right triangle is given by the formula:
A = (1/2) * base * height
In this case, the base is one of the legs (x), and the height is the other leg (12 - x). Therefore, the area can be written as:
A = (1/2) * x * (12 - x)
To find the value of x that maximizes this area, we need to find the critical points of the function A(x).
1. Take the first derivative of A(x) with respect to x:
A'(x) = (1/2) * (12 - 2x)
2. Set the derivative equal to zero and solve for x:
(1/2) * (12 - 2x) = 0
12 - 2x = 0
2x = 12
x = 6
This critical point x = 6 represents a maximum value since the second derivative (A''(x)) is negative for any x.
Thus, the leg of the triangle that maximizes its area is x = 6 cm. And the other leg can be found by subtracting x from the hypotenuse:
12 - x = 12 - 6 = 6 cm.
So, one of the legs is 6 cm, and the other leg is also 6 cm.