Find the absolute maximum and minimum values of f on set D.?
First, we find critical points inside D. f_x = 4x + 1, and f_y = 2y. Setting these equal to 0 yields (x, y) = (-1/4, 0) which is inside D. ------ Now, we check the boundary of the region. We want to find the extreme values of f(x,y) = 2x^2 + x + y^2 - 2, subject to the constraint g(x,y) = x^2 + y^2 = 4. By Lagrange Multipliers, ∇f = λ∇g. ==> <4x + 1, 2y> = λ<2x, 2y> ==> 4x + 1 = 2λx and 2y = 2λy. So, 2λxy = y(4x + 1) = 2xy ==> 4xy + y = 2xy ==> (2x + 1)y = 0 ==> x = -1/2 or y = 0. Substituting these into g yields (x, y) = (-1/2, ±√15/2), (±2, 0). ------------- Testing all of the critical points: f(-1/4, 0) = -17/8 <---Minimum f(-1/2, ±√15/2) = 7/4 f(2, 0) = 8 <---Maximum f(-2, 0) = 4 I hope this helps!
Find the absolute maximum and minimum values of f on the set D. f(x, y) = 4x + 6y - x2 - y2 + 7 D = {(x, y) |?
First, we find critical points inside D. f_x = 4 - 2x, f_y = 6 - 2y. Set these equal to 0: (x, y) = (2, 3). Note that f(2, 3) = 20. -------------------- Next, we check the boundary of D. (i) x = 0 and 0 ≤ y ≤ 5 ==> f(0, y) = 6y - y^2 + 7 = 16 - (y - 3)^2. Max = 16 at y = 3, and Min = 9 at y = 0. (ii) x = 4 and 0 ≤ y ≤ 5 ==> f(4, y) = 6y - y^2 + 7 = 16 - (y - 3)^2. Max = 16 at y = 3, and Min = 9 at y = 0. (iii) y = 0 and 0 ≤ x ≤ 4 ==> f(x, 0) = 4x - x^2 + 7 = 11 - (x - 2)^2 Max = 11 at x = 2, and Min = 7 at x = 0 or 4. (iv) y = 5 and 0 ≤ x ≤ 4 ==> f(x, 5) = 4x - x^2 + 12 = 16 - (x - 2)^2 Max = 16 at x = 2, and Min = 12 at x = 0 or 4. ------------------- Overall, the absolute maximum is 20 and the absolute minimum is 7. I hope this helps!
Find the absolute maximum and minimum values of f on the set D. f(x, y) = x^3 − 3x − y^3 + 12y + 8?
First, we find any critical points of f inside D. f_x = 3x^2 - 3, f_y = -3y^2 + 12. Setting these equal to 0 yields (x, y) = (-1, 2), (-1, -2), (1, 2), (1, -2); only (1, 2) and (-1, 2) are inside D. Note that f(1, 2) = 22 and f(-1, 2) = 26. ---------------- Next, we look extreme values of f on the boundary of D, one edge at a time (sketch it!) (i) x = 2: f(2, y) = -y^3 + 12y + 10 with y in [2, 3]. Setting f ' = -3y^2 + 12 = 0 yields y = 2 (an endpoint; ignore y = -2). Testing endpoints: f(2, 2) = 26 and f(2, 3) = 19. --- (ii) x = -2: f(-2, y) = -y^3 + 12y + 6 with y in [-2, 3]. Setting f ' = -3y^2 + 12 = 0 yields y = 2 (an endpoint; ignore y = -2). Testing endpoints: f(-2, -2) = -10 and f(-2, 3) = 15. --- (iii) y = 3: f(x, 3) = x^3 - 3x + 17 with x in [-2, 2]. Setting f ' = 3x^2 - 3 = 0 yields x = -1, 1. Testing critical and end points: f(-2, 3) = 15, f(-1, 3) = 19, f(1, 3) = 15, f(2, 3) = 19. --- (iv) y = x: f(x, x) = 9x + 8 with x in [-2, 2]. This has extreme values f(2, 2) = 17 and f(-2, -2) = -10. --------------------- Overall, the extreme values are -10 and 26. I hope this helps!
How do I find the absolute maximum and minimum values of [math]f\left(x\right)=\displaystyle\frac{x}{x^2-x+25}[/math] on the interval [math]\left[0,15\right]\subset\mathbb{R}[/math]?
Taking a derivative is when you are finding the rate of change or slope at a point. The function goes up, if it hits the end of an interval, that could be a maximum. The function goes up, and then comes down, that could also be a maximum. How do we find that maximum? On the interval that the values are increasing, what does the derivative look like? It's positive. And on the way down, the derivative is negative. If the function is continuous, then there must be a point of interest where the derivative is equal to 0. So any maximum value is going to be at a point where the derivative is equal to zero or at the end of the interval. The same is true for minimum values. f '(x) = 0Let's apply this to a different function, let's suppose:f(x) = x^2 [-2, 3]f'(x) = 2xThe only time the derivative is equal to zero, 0 = 2x, is when x=0. At x=0, x^2 = 0.At x=0.1, x^2 = 0.01At x=-0.1, x^2 = 0.01So we can see, that x=0, is a minimum. The only other interesting points is at the ends, x=-2, we get x^2 = 4, and at x=3, we get x^2 = 9So our interesting points are x=0, -2, and 3. 0^2 = 0, -2^2 = 4, 3^2=9So out of those points, we know x=0 is where our minimum, and x=3 is our maximum.
How do i find the absolute maximum and minimum values of f?
You are correct about the absolute minimum of f on D. For the maximum it is clear that x^2 + y^2 = 3. Find y^2 in terms of x, hence y^2 = 3 - x^2. Then f(x,y) = x(3 - x^2) + 3 = -x^2 + 3x + 3 such that 0 <= x <= 3. Therefore, f '(x,y) = -2x + 3 --> x = 3/2 at maximum. Thus x=3/2 and y=(27/2)^.5, and f(3/2, (27/2)^.5) = 13.125 13.125
Find the absolute max. and min. values of f on the set D ?
fx = 2x + 2y = 0 fy = 2x + 6y = 0 thus the critical point is (0,0) << this is inside region D consider other boundaries: y = 1 x = -1 y = x - 1 when y = 1 f(x) = x^2 - 2x + 3 f'(x) = 2x - 2 = 0 x = 1 thus other points to consider: (-1,1) ... (1,1) ... (2,1) when x = -1 f(y) = 1 - 2y + 3y^2 f'(y) = -2 + 6y = 0 .... y = 1/3 other critical points: (-1, 1/3) , (-1,-2) when y = x -1 f(x) = x^2 + 2x(x-1) + 3(x-1)^2 f(x) = 6x^2 - 8x + 3 f'(x) = 12x - 8 = 0 x = 2/3 thus last critical point (2/3, -1/3) f(0,0) = 0 f(-1,1) = 2 f(1,1) = 6 f(2,1) = 11 f(-1,1/3) = 2/3 f(-1,-2) = 17 f(2/3, -1/3) = 1/3 thus absolute maximum point (-1,-2,17) absolute minimum point (0,0,0) .
Find the absolute maximum and minimum values of f on the set D.?
Critical points: f_x = 3x^2 - 3, f_y = -3y^2 + 12. Setting these equal to 0 (and remaining inside D) yields the two critical points (x, y) = (1, 2), (-1, 2). ------------- Now check the boundary of D for extreme values, one edge at a time. (i) x = -2 with y in [-2, 3]: g(y) := f(-2, y) = -y^3 + 12y g'(y) = -3y^2 + 12. So, g'(y) = 0 ==> y = -2, 2. Now, check the points (-2, -2), (-2, 2), and (-2, 3). ----------- (ii) x = 2 with y in [-2, 3]: g(y) := f(2, y) = -y^3 + 12y + 4 g'(y) = -3y^2 + 12. So, g'(y) = 0 ==> y = -2, 2. Now, check the points (2, -2), (2, 2), and (2, 3). -------- (iii) y = 3 with x in [-2, 2]: g(x) := f(x, 3) = x^3 − 3x + 11. Setting g'(x) = 3x^2 - 3 = 0 yields x = -1, 1. So, check the points (-2, 3), (-1, 3), (1, 3), (2, 3). -------- (iv) (-2, 2) to (2, 2) via y = x: ==> g(x) := f(x, x) = 9x + 2 on [-2, 2]. Being linear, its extreme values occur at the edges. ----------- Finally, substitute in all of these points (critical points and those from (i) to (iv)) to determine the extreme values of f on D. I hope this helps!
How do I find absolute maximum and minimum values of a function with open intervals?
With open intervals, a continuous function is not guaranteed to have absolute extrema like with closed intervals.Your only candidates for absolute extrema are the relative extrema. You can then look at the limit of the function as you approach either endpoint. If the limit is greater than every relative maximum, there is no absolute maximum; if the limit is less than every relative minimum, there is no absolute minimum.So, for example, your function [math]f(x)=x^2-x-2[/math] has a relative minimum when [math]x=\frac 12.[/math] We know right away that there can be no absolute maximum (and indeed [math]\lim_{x \to \pm \infty}f(x)=\infty[/math]). But this relative minimum is an absolute minimum (for the same reason).
Help finding the absolute maximum and minimum values of f on the set D.?
x^2 + y^2 ≤ 3, y^2 ≤ 3 - x^2, Hence max value of y^2=(3 - x^2) now, for max value.. f(x, y) = xy^2 + 7, f(x, y) = x(3 - x^2)+7, f(x, y) = (3x - x^3)+7, f(x, y) = -x^3+3x+7 For maxima diffrntiatng wrt x.. f '(x,y)= -3x^2+3 =0, x^2=-3/-3, x^2=1, x=1. max f(x,y)=-1^3+3(1)+7 =9. y ≥ 0, Hence Min value of y^2=0. now, for Min value.. f(x, y) = xy^2 + 7, = x(0)+7, = 7. so max value of f(x,y) is 9. & min value of f(x,y) is 7.
How do I find minimum absolute value of | A * x + B * y + C |?
A strange problem. Here is my solution. It would be simpler if x=0 and y=0 were allowed.if A, B, C are all ≥ 0 the straightforward solution is |A+B+C|, i.e. x=y=1, since with any other choice of x, y the sum can be decreased decreasing either x or y.if A, B, C are all ≤ 0, same result because of the absolute value.if A = 0, B < 0, C > 0 then the minimum is obtained through the remainder R of the integer division of C by |B|, or |R-|B||, whatever is smaller. If B > 0 and C < 0, similar solution. And of course B = 0, A, C different signs is identical. There is a special case when C < |B|, since y has to be ≥1if A > 0 and B < 0 (or A < 0, B > 0) the set of all the Ax + By with x, y positive is the set of all the multiples (positive and negative) of the greater common divisor D of A and B. The approach is similar to the previous case, with B replaced by D: find the smallest remainder (in absolute value) of C by D.