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Consider the function f(x, y) = 5 – 4 sin x + y2 for 0 < x < 2p and y ∈ R. The set of critical points of f(x, y) consists of
- a)a point of local maximum and a point of local minimum
- b)a point of local maximum and a saddle point
- c)a point of local maximum, a point of local minimum and a saddle point
- d)a point of local minimum and a saddle point
Correct answer is 'D'. Can you explain this answer?
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Most Upvoted Answer
Consider the function f(x, y) = 5 – 4 sin x + y2 for 0 < x &...
Explanation:
To find the critical points of the function f(x, y), we need to find the values of x and y where the partial derivatives of f with respect to x and y are equal to zero or do not exist.
Step 1: Find the partial derivative of f with respect to x:
∂f/∂x = 4cos(x) - 0 (since the derivative of y^2 with respect to x is zero)
Step 2: Set ∂f/∂x = 0 and solve for x:
4cos(x) = 0
cos(x) = 0
The solutions to this equation are x = π/2 and x = 3π/2.
Step 3: Find the partial derivative of f with respect to y:
∂f/∂y = 0 - 2y
∂f/∂y = -2y
Step 4: Set ∂f/∂y = 0 and solve for y:
-2y = 0
y = 0
Step 5: Combine the critical points (x, y) found from Steps 2 and 4:
The critical points are (π/2, 0) and (3π/2, 0).
Step 6: Classify the critical points:
To classify the critical points, we can use the second partial derivative test.
Step 6.1: Find the second partial derivatives of f:
∂^2f/∂x^2 = -4sin(x)
∂^2f/∂y^2 = -2
∂^2f/∂x∂y = 0 (since the derivative of 4sin(x) with respect to y is zero)
Step 6.2: Evaluate the second partial derivatives at the critical points:
At (π/2, 0):
∂^2f/∂x^2 = -4sin(π/2) = -4
∂^2f/∂y^2 = -2
At (3π/2, 0):
∂^2f/∂x^2 = -4sin(3π/2) = 4
∂^2f/∂y^2 = -2
Step 6.3: Apply the second partial derivative test:
For a local maximum or minimum, the second partial derivatives must satisfy:
∂^2f/∂x^2 > 0 and (∂^2f/∂x^2)(∂^2f/∂y^2) - (∂^2f/∂x∂y)^2 > 0
At (π/2, 0):
-4 > 0
(-4)(-2) - 0^2 = 8 > 0
Therefore, (π/2, 0) is a local maximum.
At (3π/2, 0):
4 > 0
(4)(-2) - 0^2 = -8 < />
Therefore, (3π/2, 0)
To find the critical points of the function f(x, y), we need to find the values of x and y where the partial derivatives of f with respect to x and y are equal to zero or do not exist.
Step 1: Find the partial derivative of f with respect to x:
∂f/∂x = 4cos(x) - 0 (since the derivative of y^2 with respect to x is zero)
Step 2: Set ∂f/∂x = 0 and solve for x:
4cos(x) = 0
cos(x) = 0
The solutions to this equation are x = π/2 and x = 3π/2.
Step 3: Find the partial derivative of f with respect to y:
∂f/∂y = 0 - 2y
∂f/∂y = -2y
Step 4: Set ∂f/∂y = 0 and solve for y:
-2y = 0
y = 0
Step 5: Combine the critical points (x, y) found from Steps 2 and 4:
The critical points are (π/2, 0) and (3π/2, 0).
Step 6: Classify the critical points:
To classify the critical points, we can use the second partial derivative test.
Step 6.1: Find the second partial derivatives of f:
∂^2f/∂x^2 = -4sin(x)
∂^2f/∂y^2 = -2
∂^2f/∂x∂y = 0 (since the derivative of 4sin(x) with respect to y is zero)
Step 6.2: Evaluate the second partial derivatives at the critical points:
At (π/2, 0):
∂^2f/∂x^2 = -4sin(π/2) = -4
∂^2f/∂y^2 = -2
At (3π/2, 0):
∂^2f/∂x^2 = -4sin(3π/2) = 4
∂^2f/∂y^2 = -2
Step 6.3: Apply the second partial derivative test:
For a local maximum or minimum, the second partial derivatives must satisfy:
∂^2f/∂x^2 > 0 and (∂^2f/∂x^2)(∂^2f/∂y^2) - (∂^2f/∂x∂y)^2 > 0
At (π/2, 0):
-4 > 0
(-4)(-2) - 0^2 = 8 > 0
Therefore, (π/2, 0) is a local maximum.
At (3π/2, 0):
4 > 0
(4)(-2) - 0^2 = -8 < />
Therefore, (3π/2, 0)
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Community Answer
Consider the function f(x, y) = 5 – 4 sin x + y2 for 0 < x &...
Find the partial derivatives of f(x,y) wrt x and y.
fx =-4 cos x
fy =2y
equation each to zero together critical points.
fx=0 gives x =π/2, 3π/2 as 0< /><π>
fy=0 gives y=0
so critical points are (0,π/2) & (0,3π/2)
now to find local max or min we need to calculate discriminant D =fxx.fyy-(fxy)^2 So,
fxx= 4sin x
fyy= 2
fxy= 0
so D= 8 sin x and at (π/2,0), D= 8>0 and
fxx = 4>0 so there exists a local minima at (π/2,0) .Now
at (0,3π/2) D=-8 <0 i.e.="" a="" saddle="" point="" at="">
(0,3π/2). so d) is correct.
fx =-4 cos x
fy =2y
equation each to zero together critical points.
fx=0 gives x =π/2, 3π/2 as 0< /><π>
fy=0 gives y=0
so critical points are (0,π/2) & (0,3π/2)
now to find local max or min we need to calculate discriminant D =fxx.fyy-(fxy)^2 So,
fxx= 4sin x
fyy= 2
fxy= 0
so D= 8 sin x and at (π/2,0), D= 8>0 and
fxx = 4>0 so there exists a local minima at (π/2,0) .Now
at (0,3π/2) D=-8 <0 i.e.="" a="" saddle="" point="" at="">
(0,3π/2). so d) is correct.
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Consider the function f(x, y) = 5 – 4 sin x + y2 for 0 < x < 2p and y ∈ R. The set of critical points of f(x, y) consists ofa)a point of local maximum and a point of local minimumb)a point of local maximum and a saddle pointc)a point of local maximum, a point of local minimum and a saddle pointd)a point of local minimum and a saddle pointCorrect answer is 'D'. Can you explain this answer?
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