Lagrange Multiplier Method: Maximal Perimeter Rectangles

Practice with Lagrange Multiplier Problems. Worked example recitation video Solving multivariable optimization problems using the method of Lagrange multiplierscan be challenging, but we have a nice problem here for you that can be solved thatway. In this problem we've got an ellipse — the ellipse with equation x squared plus4y squared equals 4 — and we want to inscribe a rectangle in it. Actually, a rectanglewhose edges are parallel to the axes. So we want to inscribe a rectangle in thisellipse, and find among all such rectangles, find the one with the largest perimeter. Let's get started on it. One thing we need to start is figuring out a way to describethese rectangles, and then in a way that will let us describe their perimeter: writedown what their perimeter is. So a natural way to do that is to call thisupper-right-hand corner of the rectangle—to call it the point (x, y). So (x, y) is goingto be that upper-right-hand corner of the rectangle, and it's going to be ranging overthe region from this topmost point on the ellipse down to this rightmost point on theellipse—on this quarter arc of the ellipse. The perimeter of a rectangle is the sum of its horizontal and vertical lengths. So if thepoint (x, y) is inside the rectangle, then we need to figure out what the maximumperimeter is from that point. The perimeter will be a function of x and y, so x is thisdistance. So the length of the horizontal edge of the rectangle is 2x. We've got two ofthose, so it's 4x from the horizontal sides. And then this height is y, so the length ofthe vertical side of the rectangle is 2y. So then the perimeter is going to be 4x plus4y. Now that's our objective function that we're trying to find the maximum of. We also have the constraint function, g, which is x squared plus 4y squared, and theconstraint is that the g is equal to 4. So we have the objective function, p (x, y) andwe have this constraint function, g. And so we want to write down some equationsusing Lagrange multipliers whose solutions will correspond to the possible maximumpoints of P.If we assume that the gradients of P and g are parallel, then we will have P sub x =lambda × g sub x and P sub y = lambda × g sub y for some value of lambda. Weneed to find a value of lambda that makes this true. And then also our third equation is the constraint equation that g is equal to 4. Sowhat does P sub x equal lambda g x translate to in our case? We just need to plug inthe values of P sub x and g x into this equation. In our case, P sub x is the firstpartial derivative of 4x plus 4y; so that's just 4. And g sub x is--we'll take the partialderivative with respect to x of x squared plus 4y squared and that's 2x. So 4 is equalto lambda times 2x.We can find the partial derivative of the function P with respect to y by taking thederivative of each term in P with respect to x and substituting that into the formula for

the partial derivative of a product. The resulting expression is 4y. Since lambda isequal to 8y, we can conclude that lambda equals 4. We will solve the system of equations by isolating lambda between the first equationand second equation. We can see that x has to be exactly four times as large as yfor both of these equations to be true at the same time. So we need x to be equal to4y. Now, what is x? Well, we've solved for x in terms of y so that x is equal to 4 oversquare root of 5. So Lagrange multipliers--when we use the method of Lagrangemultipliers - we get one possible point at which we have to check to be themaximum. But remember that when you're using Lagrange multipliers, you also haveto worry about the boundary of the region that you're interested in.

On our graph, the point x, y, moves along an arc connecting the topmost point of theellipse to the rightmost point of the ellipse. We also have to calculate the perimeterwhen x is at its highest point in the graph and when x is at its lowest point.In the cases of x = 0 and y = 1, the function is degenerate, and when x = 0 and y = 1it looks like two copies of the vertical line, the minor axis. When x = 2 and y = 0, itjust looks like the major axis - just that horizontal line. But we still must checkwhether our function has a maximum, and what it is. So we must compute theobjective function value at this point, as well as at those endpoints.We need to compare the values of P at the points given by Lagrange multipliers,which are 4 times the square root of 5 and 8, with those along the boundary arc,which are 4 and 8. Since 4 times the square root of 5 is larger than both 4 and 8, wecan conclude that it is the maximum possible value of P in this region. This is thelargest, so this is actually the maximum value. So the maximum perimeter is 4(sqrt5) when the rectangle has its upper rightmost vertex at the point (4 sqrt 5, 1 sqrt 5).So our rectangle's maximal perimeter is 4(sqrt 5), and that occurs when the upperrightmost vertex is at the point (4 sqrt 5, 1 sqrt 5). To recap, we wanted to apply the method of Lagrange multipliers to this problem. Sowe chose to keep track of our rectangles by their upper-right-hand corner and thenthat gave us the perimeter was 4x plus 4y. That was our objective function. And theconstraint was that that upper-right-hand corner actually had to lie on the ellipse. Sowe set the gradients of the two functions parallel and solved the system of equations.There's some constant multiple lambda that appears. So we set the gradients to beparallel to each other, and the constraint equation to hold, and we solve those threeequations simultaneously for x and y. And those equations gave us one point that wehad to check to be the maximum, and we also needed to check points on theboundary of the region in question.