find the largest possible value of s=2x+y if x and y are side lengths in a right triangle whose hypotenuse is 5 units long.
My answer :
(5)^2 = x^2 + y^2
y^2 = 5 - x^2
y = 5-x^2
Substituting y to s,
S = 2x + y = 2x + 5-x^2
To find the largest possible value,
ds/dt = ….? I confuse to complete this, please help me
s = 2x + y = 2x + sqrt(5-x^2)
so s = 2x + √(5 - x^2)
= 2x + (5-x^2)^(1/2)
why did you try to find ds/dt ?
That would be a rate, and there is not mention of any rate or rates of change.
ds/dx = 2 + (1/2)(5 - x^2)^(-1/2) (-2x)
= 2 - x/√(5-x^2)
= 0 for a max/min of s
2 = x/√5-x^2)
2√(5-x^2) = x
square both sides
4(5-x^2) = x^2
20 =5x^2
x^2 = 4
x = ±2 , but x is a side, so it can't be negative
x = 2, then y = √(5-4) = 1
so the largest possible value of
s = 2(2) + 1 = 5
thanks for your help Reiny :)
To find the largest possible value of S = 2x + y, we can take the derivative of S with respect to x and set it equal to zero. This will help us find the critical points where the maximum value occurs.
Taking the derivative of S = 2x + y with respect to x:
dS/dx = 2 + dy/dx
Now, we need to find dy/dx. From the equation y = √(5 - x^2), we can solve for dy/dx:
dy/dx = d/dx[√(5 - x^2)]
= (1/2)(5 - x^2)^(-1/2) (-2x)
= -x / √(5 - x^2)
Substituting this back into the derivative of S:
dS/dx = 2 - x / √(5 - x^2)
To find the critical points, we set dS/dx equal to zero:
2 - x / √(5 - x^2) = 0
Solving for x:
2 = x / √(5 - x^2)
Simplifying further:
4 = x^2 / (5 - x^2)
4(5 - x^2) = x^2
20 - 4x^2 = x^2
5x^2 = 20
x^2 = 4
x = ±√4
x = ±2
Since we are looking for the largest possible value, we consider the positive value for x, which is x = 2.
Now, substituting this value of x into the equation y = √(5 - x^2):
y = √(5 - 2^2) = √(5 - 4) = √1 = 1
We have found the values of x and y for the largest possible value of s. Now let's substitute these values into the original equation:
s = 2x + y = 2(2) + 1 = 4 + 1 = 5
Therefore, the largest possible value of s = 2x + y, where x and y are side lengths in a right triangle with a hypotenuse of √5 units long, is 5.