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This critical number calculator determines those points on which the function .Questions & answersStack ExchangeQuestionCalculate the partial derivative, local minima/maxima, and saddle points.Answer · 1 votePartial derivatives are calculated by regarding the function as a function in only one argument and considering the other variables as constants.So you obtain fx(x,y)=ddx(yx−2x−1−2/y)=y+2x−2.Saddle points and local maxima/minima are always at places where both derivatives vanish simultaneously. To decide which is which, you have to look at the second derivatives:• fx(x0,y0)>0,fy(x0,y0)>0: (x0,y0) is local minimum• fx(x0,y0)<0,fy(x0,y0)<0: (x0,y0) is local maximum• fx(x0,y0)>0,fy(x0,y0)<0 orfx(x0,y0)<0,fy(x0,y0>0: (x0,y0) is a saddle point.MoreMathWorksQuestionCalculating max, min, and saddle pointsAnswer · 1 votesyms x ya = ((y.*x^3)+(x.*y^3)-(1.69.*x.*y));eqA_partial_diff_x = diff(a,x)eqA_partial_diff_y = diff(a,y)M = solve(eqA_partial_diff_x,eqA_partial_diff_y)N = [M.x,M.y][m,n]=size(N)eqA_double_partial_diff_x = diff(eqA_partial_diff_x,x)eqA_double_partial_diff_y = diff(eqA_partial_diff_y,y)eqA_double_partial_diff_y_then_x = diff(eqA_partial_diff_y,x)x=N(:,1)y=N(:,2)eqA_double_partial_diff_x=subs(eqA_double_partial_diff_x)eqA_double_partial_diff_y =subs(eqA_double_partial_diff_y)eqA_double_partial_diff_y_then_x=subs(eqA_double_partial_diff_y_then_x)%Determinant partsyms x ya = ((y.*x^3)+(x.*y^3)-(1.69.*x.*y));eqA_fx = diff(a,x)eqA_fy = diff(a,y)M = solve(eqA_fx,eqA_fy)N = [M.x,M.y]eqA_fxx = diff(eqA_fx,x)eqA_fyy = diff(eqA_fy,y)eqA_fyx = diff(eqA_fy,x)D = eqA_fxx.* eqA_fyy - eqA_fyx^2subs(D,{x,y},{N(:,1),N(:,2)})|*Remark: You don’t need loop to read the values one by one instead you can substitute all the values at once*|MoreCheggQuestionFind the local maximum and minimum values and saddle point(s) of the function. You are encouraged to use a calculator or computer to graph the function with a domain and viewpoint that reveals all the important aspects of the function. (Enter your answers as comma-separated lists. If an answer does not exist, enter DNE.) f(x,y)=(x−y)(25−xy) local maximumAnswer · 0 votesTo find the local maximum and minimum values and saddle points of the function we first calculate the.MoreWyzantQuestionFind Critical Point and classify as rel. min, rel, max, or saddle point.Answer · 1 voteTo find critical points, we first determine the partial derivatives of the function f(x,y) = 2y2 - x2 + 6x - 4:Partial derivative of f w/ respect to x:fx = -2x + 6Partial derivative of f w/ respect to y:fy = 4yNext, we determine any critical points by finding points (x,y) where both partial derivatives are 0:-2x + 6 = 04y = 0We can solve this system of equations:-2x + 6 = 06 = 2x (add 2x to both sides)x = 3 (divide both sides by 2)and4y = 0y = 0 (divide both sides by 4)So our only critical points is at x = 3 and y = 0: (3,0).To determine whether this is a rel. min, max or saddle point: let's use the second derivative test. Find the second derivatives of f:fxx = -2fyy = 4fxy = 0Now we apply the formula: D(x, y) = fxx ⋅ fyy - (fxy)2.D(0,0) = -2 ⋅ 4 - (0)2D(0,0) = -8The partial derivative formula tells us that if D is negative, then the critical point is a saddle point.Thus our solution is:Critical point(s) f(x,y) = 2y2 - x2 + 6x - 4:(3,0), a saddle pointMoreWyzantQuestionLocal Max, Min, Saddle PointAnswer · 1 voteOne thing we know about the local minima and maxima of a function of two variables is that they occur at critical points of our function. These critical points are where the gradient of our function, ∇f(x,y), satisfies, ∇f(x,y) = (fx(x,y), fy(x,y)) = (0,0). So every pair of points (x,y) that satisfy this equation is a critical point. It's important to note that not every critical point is a local min/max. We'll need some additional information to determine that.A hint to make sure your heading in the right direction: In this case, you should have two values of x that satisfy fx(x,y) = 0 and two values of y that satisfy fy(x,y) = 0, meaning you have four critical points in total.Second, to determine whether each point is a local max/min or saddle point, we need to implement the second derivative test. The test tells us precisely that if we letD(x0,y0) = fxx(x0,y0)fyy(x0,y0) - (fxy(x0,y0))2for a critical point (x0,y0)if D(x0,y0) > 0 and fxx(x0,y0) >0, f(x0,y0) is a local minif D(x0,y0) >…MoreNumeradeQuestionFind the local maximum and minimum values and saddle point(s) of the function. You are encouraged to use a calculator or computer to graph the function with a domain and viewpoint that reveals all the important aspects of the function. f(x, y)=x^2+x y+y^2+yAnswer · 3 votesWe have the function f of x y is equal to x, squared plus x y plus y squared plus y, so we are going to find where it has a maximum or a minimum. It says we're encouraged to use graphing calculators to figure out what this looks. Like so just to give you a an ideal kind of just looks like this, like some kind of bole or something so we should be finding this point down here. This is- and we should be finding this point down here so or to do that. We need to differentiate with respect to x and we get 2 x plus y. We need to differentiate with respect to y and we get x, plus 2 y plus 1, and we're going to set these each equal to 0. Then we can use system of equations to solve so for this top equation. 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