Posted by **rolan** on Monday, August 13, 2012 at 12:31am.

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

- math -
**rolan**, Monday, August 13, 2012 at 12:38am
s = 2x + y = 2x + sqrt(5-x^2)

- math -
**Reiny**, Monday, August 13, 2012 at 9:16am
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

- math -
**rolan**, Monday, August 13, 2012 at 11:01am
thanks for your help Reiny :)

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