Find the absolute maximum and minimum values of f(w,y) = lx + 2y within the domain 2? + y? <2.Absolute minimum value of f(w,y) is 2. Absolute maximum value of f(...

Find the absolute maximum and minimum values of cach function, subject to the given constraints. $-g(x, y)=x^{2}+2 y^{2},-1 \leq x \leq 1$ and $-1 \leq y \leq 2$

Question number one. We need to find a little Maxima and minima off this particular function So forceful. Differentiate with respect to X. Let's noted by f and with a subscript exit means that we're differentiating with respect to X treating while the constant so FX will be different station of X squares to x x y will just to be. Why? Because why is a constant on a wire square zero and off? Negative wires also zero. So we have two x plus y as the terrible, terrible respective eggs like where is the deliberate, the derivative? With respect, Why would be zero? Creating X is a constant This this would be excess would be to buy. This would be minus one, so we have X plus two by minus one. Now, what we do is really equate FX +20 and f y +20 to get, uh, critical numbers. So we have fxe zero. This means that two x plus y zero This means that why is negative two works? It's called Equation one, while for F from every we have X plus two by minus one, this is equated to zero. We already have the value of y as negative toe works or replacing Why bynegative to work. We have negative three Xs one, it needs the excess negative one or three. If X is negative one or three, then why would be negative two times x from this equation? So that will be to over three. So a point off interest would be negative. One or three and two or three. Uh, while now we need to verify with others, has a maximum rate of animals. So that's where F double dash F X X comes in derivative find, uh, derivatives again with respect to work. So that will just be too. Then we find out why. Why? So we decorate this, uh, regrettable. This with respect to y will just be too on. Then we get we find x y So ex wife from here will just be one. Now we find me which is f x six times f y y minus f x y square. It comes out as four minus one, which is three. So since the d is positive on F X X is also greater than zero. It means that we have ah, relative minimum. This implies that we have the relative minimum at the critical point. Let's rewrite this little minima at minus one or three on to over three. And the value off F X Y at this point would be minus one or three whole square. But ex wife. So we have X. Why? Plus why? Square minus y. So this is 119 minus 2/9 plus four or nine minus 2/3. So this comes out as, uh, 3/9, minus two or three, which is nothing but minus from one or three. So this is the relative minimum. While this is the point where we have the relative

For this problem, we are asked to find the absolute maximum and absolute minimum values of the function H of x, Y equals x squared plus y squared minus four, X minus two, Y plus one. Subject to the constraints that x and Y are greater than or equal to zero and X plus two Y is less than or equal to five. So the first thing that we want to do here is look for critical points of our function. So we do that by trying to set our gradients to equal zero. So we have two x minus four equals zero and two, y minus two equals zero. So we can see that that would lead to x equals two, Y equals one and we can see that X plus two, Y would be two plus just yeah, it would be two plus two. Less than or equal to five is satisfied. So we do have a valid critical point here. So evaluating our function at that point we want age of 21 is going to be four plus one minus um yeah minus four times two, so minus eight minus two. So let's see here uh it would be five minus 10, so we'll have a value of negative five. So we it seems like that's probably going to be our minimum then we want to so essentially solve the constraint or solve them trying to figure out how to phrase this essentially for the second part we will apply the method of lagrange multipliers treating that constraint equation as an exact or as an equality rather than an inequality. So we want to have our gradients of our function equal lambda times the gradients of our constraint. So we have two x minus four equals lambda. And then that would mean also we would have two y minus two, little equal to lambda. And lastly we have then that X plus two, Y As two Equal 5. So we can easily see from the first or from that last equation that X equals 5 -2. Y. You can then substitute that into our first equation there. So we'll have that 10 -4 Y -4 equals lambda. So that's supposed to be a lambda. So we have then that λ is going to equal 6 -4. Y. Then substitute that into our second equation. So we'll have that 12 -8. Y Equals two Y -2. So adding to to both sides and adding eight white both sides, we'll get that 10 Y is going to equal um Let's see here. Yeah it's going to be that 10 Y will equal um 14. So why is going to equal 14/10 Which we can reduce down to 7/5. One second here. All right. So yeah we have Y equals 7/5 which then means that X is going to equal 5 -14/5 or 10 -14/5 or 10. 10/5. No no never mind. 25 minus 14/5. So that is going to be the same thing as 20 or 25 -10 -4. So 25 -10, maybe 15 -4, would be 11/5 for X.

For this problem, we are asked to find the minimum of the function F. Of X. Y equals X times Why is subject to the constraint that X squared plus Y squared equals nine. So what we want to do is we have that our G of X. Y. Our constraint function can be expressed as X squared plus Y squared minus nine. And we want to then find lambda X. And Y. Such that the gradient of F equals lambda times the gradient of G. So taking the derivatives. That means that we need Why to equal? Let's see here. That would be we want Y to equal two times lambda X to lambda X. Here. And then with respect to why we'll get that X must be equal to lambda. Why? Which in turn, we know that why equals 2 λ X. So you can plug that in. We'd have that X equals And is misbehaving X would equal two times λ Times 2 λ X. So we have that. We can actually write this as being first of all, that one is going to equal four times lambda squared. So λ is going to equal plus or -2 or actually plus or minus 1/2 Because we divide through both sides by four square root. All right. So, we have that then. And we also have that Y equals two times lambda times X. Now we need to figure out what X is. We can do that. Bye. We know that. What why is in terms of X. And we know what lambda is. So we can plug that into our expert. Plus Y squared equals nine equation there. So we get that X squared plus four lambda squared X squared must equal nine for lambda square two, X squared is going to just go to X squared because we know that land is plus or minus one half. So we get that two X squared equals nine or in turn that X squared equals nine over to. So X equals plus or minus route or plus or minus three over root two, Say 1 2nd here. All right. Yeah. So we have X equals plus or minus three over root two, which then means that why is going to equal plus or minus? Um two times 1/2. So yeah, it's going to be equal to then plus or minus. Um Actually, it's just going to shake out to being plus or minus X. So plus or minus three over root two as well. So what we need to do then is evaluate our function at each one of those different possible points. Or we could just recognize that since we have we are dealing with F of X. Y. That to get the maximum we'll plug in the largest values. So we would have f of three over root two Times three Over Root two. So that's going to be nine over root two times two. So 9/2 is the maximum and then the minimum evaluated at negative three over root two, Positive three over route to being careful because if we multiply the two negatives together, we would end up with a positive value again. But if we take a negative and a positive then we'll get negative 9/2. And that also equals the value of the function at the 0.3 over root two times -3 Over Root two. And I'll note that that value was positive three over route to positive three over root two. It's going to be the same thing as the value of the function at negative three over root two, negative 3/2.

This problem, we are asked to find the absolute maximum and minimum values of the function f of x y equals x squared plus y squared minus two, x minus two. Y. Subject to the constraints that X is between zero and four and y is between zero and three. So our first step is that we want to find the critical points. So we can do that by trying to find where the gradient of our function equals zero. So taking partial derivative with respect to X. We want two x minus two to equal zero, which clearly means that x will equal one. And we also have that 2, -2 would equal zero. Which means that why would equal one? And evaluating our function at the .1, 1 we get a value of -2 as our minimum. Then to find our essentially our boundary points where we will have the maximum since we have no other critical points, the maximum must be found at the boundaries. So now we set our we set our boundaries to be essentially equality, ease rather than um rather than being inequalities. So specifically what we would do actually one second here. Yeah, we can see then that if we evaluate our function at the .4, 3, Half of four or 3 We get a value of 11. And if we evaluate our function at the zero, we get a value of just zero which is actually greater than the minimum. So we can see them that our maximum is going to be that .11 at that point where it has a valuable.