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
3 answers
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
= 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 :)
1 answer