How do I find maximum and minimum value of a function?
There are Various methods in order to find maximum or minimum value of a function. One of the conventional methods is:Find the derivative of the function and equate it to zero.Find the roots of the differentiated equation.Do double differentiation of original function and substitute the values of roots in the 2nd differentiated expression.If the value comes out to be negative, At the particular value of the root Maximum occurs. Then substitute the value in original expression to get Maximum of the function.If the value of double derivative after substituting the root is positive, Minimum occurs. Then substitute the value in original equation to get Minimum value of the function.If the Second derivative is Zero: Then go for higher derivatives of the function & substitute the value of the root in the nth order derivative expression. If it's positive it would give the Maximum of the function at the particular root.Hope the answer Helps.
Find the local maximum and minimum values and saddle point(s) of the function. If you have three dimensional g?
The critical points are when df/dx = 0 and df/dy = 0 ; (i) df/dx = 6x² + y² + 10x = 0 (ii) df/dy = 2xy + 2y = 0 ⟹ 2y( x + 1) = 0 with solutions x = - 1 or y = 0 ; Substitute in (i) : x = -1 ⟹ 6 + y² - 10 = 0 ⟹ y² = 4 ⟹ y = ± 2 ; y = 0 ⟹ 6x² + 10x = 0 ⟹ x = 0 or x = - 5/3 : So critical points are (-1,-2) ; (-1,2) ; (0 ,0) ; ( -5/3, 0) For determining the nature of the critical points ( x₀ , y₀ ), evaluate : D(x₀,y₀) = d²f/dx² * d²f/dy² - ( d²f/dxdy )² in the critical points , this way : if D > 0 and d²f/dx² in (x₀,y₀) > 0 , then f(x₀,y₀) is a relative minimum value ; if D > 0 and d²f/dx² in (x₀,y₀) < 0 , then f(x₀,y₀) is a relative maximum value ; if D < 0 , then f(x₀,y₀) is a saddle point. d²f/dx² = 12x + 10 d²f/dy² = 2x + 2 d²f/dxdy = 2y I see that max and min points are at (0 ,0) ; ( -5/3, 0) ; and saddle points at (-1,-2) ; (-1,2) You do the evaluation. As for graphing , maybe this : http://www.wolframalpha.com/input/?i=f+%... or, if you're on Windows , try this at Microsoft ( Microsoft Mathematics ) : http://www.microsoft.com/downloads/en/de...
Finding local maximum and minimum values and saddle point or points of the function f(x,y) = 2x³ + xy² + 5x²?
First compute the partial derivatives. f_x = 6x² + y² + 10x, f_y = 2xy + 2y. Set them equal to 0: 6x² + y² + 10x = 0 2xy + 2y = 2y(x + 1) = 0 ==> y = 0 or x = -1. (i) If y = 0, then 6x² + 10x = 0 ==> x = 0 or -5/3. (ii) If x = -1, then 6 + y² - 10 = 0 ==> y = 2 or -2. This yields four critical points (0,0), (-5/3, 0), (-1,2), (-1,-2). ------------------ Next, we use the second derivative test to classify them. f_xx = 12x + 10, f_yy = 2x + 2, f_xy = 2y. So, D = (f_xx)(f_yy) - (f_xy)^2 = (12x + 10)(2x + 2) - 4y^2. (i) At (0,0), D = 20 > 0 and f_xx = 10 > 0. ==> Local minimum, with value f(0,0) = 0. (ii) At (-5/3,0), D = 40/3 > 0 and f_xx = -20 < 0. ==> Local maximum, with value f(-5/3,0) = -250/27 + 125/9 = 125/27. (iii) At (-1,2), D = -16 < 0 ==> Saddle Point. (iv) At (-1,-2), D = -16 < 0 ==> Saddle Point. I hope this helps!
Find the local maximum and minimum values and saddle point(s) of the functions.?
I have three different problems. I got an answer for the first one which was (-6,-6) was the critical point, D=3>0 therefore f(a,b) is a local maximum. Please tell me if I'm right! Who ever is able to answer these questions will get 10 points! 1. f(x,y) = xy -2x -2y -x^2 -y^2 2. f(x,y) = xe^(-2x^2 - 2y^2) 3. f(x,y) = xy (1 -x -y) THANK YOU SO MUCH! I really appreciate your help!
Finding local max, min, and saddle points of the function?
First, compute the first order partial derivatives. f_x = -10x e^y, f_y = 5e^y (y^2 - x^2) + 5e^y (2y) = 5e^y (y^2 + 2y - x^2). Set them equal to 0 to solve for x and y: The first equation tells us that x = 0. ==> y^2 + 2y - 0 = 0 ==> y = 0 or -2. So, we have two critical points (0,0) and (0, -2). To classify these, take the second order partial derivatives. f_xx = -10 e^y, f_xy = -10x e^y f_yy = 5e^y (y^2 + 2y - x^2) + 5e^y (2y + 2) = 5e^y (y^2 + 4y - x^2 + 2). When (x,y) = (0,0): D = (f_xx)(f_yy) - (f_xy)^2 = -10 * 10 - 0^2 < 0 So, this is a saddle point. When (x,y) = (0,-2): D = (f_xx)(f_yy) - (f_xy)^2 = -10e^(-2) * -10e^(-2) - 0^2 >0 and f_xx < 0. So, this is a local maximum. I hope this helps!
CALC III Local Min/Max/Saddle Point question!?
1) First, find the critical points by setting the first partials equal to 0. f_x = 2y sin x, and f_y = 2y - 2 cos x. Note that 2y sin x = 0 ==> y = 0 or sin x = 0. (i) If y = 0, then 2 * 0 - 2 cos x = 0 ==> x = π/2, 3π/2 since -1 ≤ x ≤ 7. (ii) If sin x = 0, then x = 0, π, 2π If x = 0, then 2y - 2 cos 0 = 0 ==> y = 1 If x = π, then 2y - 2 cos π = 0 ==> y = -1 If x = 2π, then 2y - 2 cos 2π = 0 ==> y = 1 --------------------- So, we have five critical points (π/2, 0), (3π/2, 0), (0, 1), (π, -1), (2π, 1). We classify these with the second order partials. f_xx = 2y cos x f_xy = 2 sin x f_yy = 2. D = (f_xx)(f_yy) - (f_xy)^2 = 4y cos x - 4 sin^2(x). For (π/2, 0), (3π/2, 0), note that D = 0. So, these are saddle points. For (0, 1), (π, -1), (2π, 1), note that D = 4 > 0 and f_xx = 2 > 0 So, these are minima. ----------------------------------- 2) f_x = cos x + cos(x + y) f_y = cos y + cos(x + y). Setting these equal to 0 and subtracting, we get cos x = cos y. We have two cases: (i) If y = x: cos x + cos(2x) = 0 2cos^2(x) + cos x - 1 = 0 (2 cos x - 1)(cos x + 1) = 0 ==> cos x = 1/2 or -1 ==> x = π/3, 5π/3 or π. ---- (ii) If y = -x: cos x + cos 0 = 0 ==> cos x = -1. ==> x = π. --------------------------- Thus, we have 3 critical points (that I could find): (π/3, π/3), (5π/3, 5π/3), (π, π). Taking second order partials: f_xx = -sin x - sin(x + y) f_xy = -sin(x + y) f_yy = -sin y - sin(x + y). D = (sin x + sin(x + y))(sin y + sin(x + y)) - sin^2(x + y) For (π/3, π/3), we have D > 0 and f_xx < 0. So, this is a local maximum. For (5π/3, 5π/3), we have D > 0 and f_xx > 0. So, this is a local minimum. For (π, π), we have D = 0. So, the second derivative test is inconclusive. It is easy to check that there are values both bigger and less than f(π, π) = 5 in the vicinity of (π, π). So, this is a saddle point. ------------------------- Note: I'm curious to see what the two other critical points are... I hope this helps!
How to find local maximum, minimum, and saddle points?
So far, so good. The critical points occur where the both partial derivatives are zero. You may wish to derive the second partial derivatives to analyze the nature of the critical points. The point is a maximum if and only if it is a maximum with respect to both variables. Likewise, it is a minimum if and only if it is a minimum with respect to both. If the critical point is a maximum with respect to one variable, but a minimum with respect to the other, then it is a saddle. Also, if the critical point is a point of inflection with respect to either or both variables, then it is a saddle of a different kind.
What is the maximum and minimum value of sin(x)+sin(y)+cos(x+y)?
This isn’t too hard.Look at it this way, required expression= [math]\sin x + \sin y + \cos (x+y)[/math]=[math]2\sin(\frac{x+y}{2})\cos(\frac{x-y}{2}) + 1 - 2\sin^2(\frac{x+y}{2})[/math]=[math]2\sin(\frac{x+y}{2}).[\cos(\frac{x-y}{2}) - \sin(\frac{x+y}{2})] + 1[/math]Since [math]\sin(\frac{x+y}{2})[/math] and [math]\cos(\frac{x-y}{2})[/math] are independent of each other (except that they are bounded.), lets take,P = [math]\sin(\frac{x+y}{2})[/math] , and Q = [math]\cos(\frac{x-y}{2})[/math]So, required expression= [math]2P.(Q - P) + 1[/math]= 2.[math][\frac{Q^2}{4}-(P-Q/2)^2][/math] + 1In order to maximise the above, Maximize [math]\frac{Q^2}{4}[/math] and minimize the (independent) [math](P-Q/2)^2[/math]Setting Q = 1 and the square term as 0, we have ,Expression max = [math]\dfrac{3}{2}[/math]Since the graph is an inverse parabola, it gets minimized at one of the end points,I can think of three possible endpoints here for the expression,one is P = Q = 1.other is P = 1, Q = -1next is P = -1 , Q = 1.Putting P = Q = 1, gives, expression = 1.Putting P = 1 , Q = -1, gives, expression = -3.Putting P = -1, Q = 1, gives , expression = -3.Thus, minimized value of expression = -3.
What is the saddle point of this function? I found the max and min to be DNE?
f(x, y) = 4y cos(x), 0 ≤ x ≤ 2π I can't type letter subscripts on here, so here's the notation I'm using: fx = partial derivative with respect to x fy = partial derivative with respect to y fxx = second partial derivative with respect to x etc. fx = -4y sin(x) fy = 4 cos(x) fx = 0 when x = 0, x = π, x = 2π or y = 0 None of the x-values above cause fy to equal zero, so discard those. fy = 0 when x = π/2 and x = 3π/2 Critical points are (π/2 , 0) and (3π/2 , 0) fxx = -4y cos(x) fxy = -4 sin(x) fyy = 0 D = (fxx)(fyy) - (fxy)² = -16sin²(x) At critical point (π/2 , 0) , D = -16sin²(π/2) = -16 < 0, so (π/2 , 0, 0) is a saddle point. At critical point (3π/2 , 0) , D = -16sin²(3π/2) = -16 < 0, so (3π/2 , 0, 0) is a saddle point. ~~~~~~ Here's the best angle I could get for a graph of the surface: http://s1164.photobucket.com/user/iago9/...