One leg of an isosceles right triangle measures 5 inches. Rounded to the nearest tenth, what is the approximate length of
1 answer:
You might be interested in
Answer:
Maximum: ((1,-2,5) ; 30)
Minimum: ((-1,2,-5) ; -30)
Step-by-step explanation:
We have the function f(x,y,z) = x - 2y + 5z, with the constraint g(x,y,z) = 30, with g(x,y,z) = x²+y²+z². The Lagrange multipliers Theorem states that, the points (xo,yo,zo) of the sphere where the function takes its extreme values should satisfy this equation:
grad(f) (xo,yo,zo) = λ * grad(g) (xo,yo,zo)
for a certain real number λ. The gradient of f evaluated on a point (x,y,z) has in its coordinates the values of the partial derivates of f evaluated on (x,y,z). The partial derivates can be calculated by taking the derivate of the function by the respective variable, treating the other variables as if they were constants.
Thus, for example, fx (x,y,z) = d/dx x-2y+5z = 1, because we treat -2y and 5z as constant expressions, and the partial derivate on those terms is therefore 0. We calculate the partial derivates of both f and g
- fx(x,y,z) = 1
- fy(x,y,z) = -2
- fz(x,y,z) = 5
- gx(x,y,z) = 2x (remember that y² and z² are treated as constants)
- gy(x,y,z) = 2y
- gz(x,y,z) = 2z
Thus, for a critical point (x,y,z) we have this restrictions:
- 1 = λ 2x
- -2 = λ 2y
- 5 = λ 2z
- x²+y²+z² = 30
The last equation is just the constraint given by g, that (x,y,z) should verify.
We can put every variable in function of λ, and we obtain the following equations.
- x = 1/2λ
- y = -2/2λ = -1/λ
- z = 5/2λ
Now, we replace those values with the constraint, obtaining
(1/2λ)² + (-1/λ)²+(5/2λ)² = 30
Developing the squares and taking 1/λ² as common factor, we obtain
(1/λ²) * (1/4 + 1 + 25/4) = (1/λ²) * 30/4 = 30
Hence, λ² = 1/4, or, equivalently,
If then 1/λ is 2, and therefore
- x = 1
- y = -2
- z = 5
and f(x,y,z) = f(1,-2,5) = 1 -2 * (-2) + 5*5 = 30
If then 1/λ is -2, and we have
- x = -1
- y = 2
- z = -5
and f(x,y,z) = f(-1,2,-5) = -1 -2*2 + 5*(-5) = -30.
Since the extreme values can be reached only within those two points, we conclude that the maximun value of f in the sphere takes place on ((1,-2,5) ; 30), and the minimun value takes place on ((-1,2,-5) ; -30).