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1) Minimize K = x? 4y2 subject to x(Give EXACT VALUES)The minimum valueand it occurs whenand y2) A decorative box will be made; The top and bottom (base) will be sq...

Question

1) Minimize K = x? 4y2 subject to x(Give EXACT VALUES)The minimum valueand it occurs whenand y2) A decorative box will be made; The top and bottom (base) will be square _ The material needed t0 make the base will cost S0.25 per square foot; while the material to make the top and the sides will cost SO.15 per square foot: The volume of cubic feet: Minimize the cost: Let be the width of the base in feet; and let _ be the box must have height in feet: Round your answers tO_ decimal placesWrite the

1) Minimize K = x? 4y2 subject to x (Give EXACT VALUES) The minimum value and it occurs when and y 2) A decorative box will be made; The top and bottom (base) will be square _ The material needed t0 make the base will cost S0.25 per square foot; while the material to make the top and the sides will cost SO.15 per square foot: The volume of cubic feet: Minimize the cost: Let be the width of the base in feet; and let _ be the box must have height in feet: Round your answers tO_ decimal places Write the objective function; B) Write the constraint: The minimum cost of the box and it occurs when und



Answers

The material for a closed rectangular box costs $\$ 2$ per square foot for the top and $\$ 1$ per square foot for the sides and bottom. Using Lagrange multipliers, find the dimensions for which the volume of the box is 12 cubic feet and the cost of the materials is minimized. [Refer to Fig. 4(a)] The cost will be $3 x y+2 x z+2 y z.$

We're trying to optimize the cost of making this rectangular box. And so I have the box and I said we have A. B. And L. As our dimensions of this spectacular box. They say that the sides cost $2 each, No $2 per square foot. Um so whatever we're making the sides out of, we're making them out of something that costs $2 per square foot. So the cost of the sides is $2 times. Um We got to A plus B. Pleasant to eight times +288 times B. Plus two eight times L. So those are the four sides, The area of the four sides. And then we have $2 per square foot. And then the lessee here was, what did they say that It was the top costs $1 per square foot, the material there, so that's $1 time. Um As B. Times L. And the bottom costs $4 per square foot. So that's four times B. Times L. And we have a volume, we need the volume to be 12. Part 10. 10 square feet are cubic feet. And So we have our augmented function. We have our costs that were trying to optimize and given our λ times are constraint with the -10. Now we have A B. C, A. B. N. L. And lambda as variable. So really respect A. We get this And I guess I didn't explicitly set that equal to zero. So that's going to equal zero and then we expect to be set that equal zero. Specter L. Set that equal zero and The expected lamb that we just get our constraint that the volume has to be 10 so we can then back substitute and everything. And so we have four equations of foreign loans. We can solve those four equations in terms of for those foreign loans. And what we get is That a. is 5/2ves and B. And L. Of both too. So we've got basically a square base and the height is a little higher than it is. The height is a little higher than the sides of the base And land equals -4. And so if we plug this back into here, we find out that we have a cost of $60 and we can also you know solve this by solving for what did I solve for here. Look at my notes here I forgot to label this plot. Ah Let's see here. Find it here. Getting closer there we are. I solved for al. So I eliminated al. So this is F. Bar and this is A. And B. I can't remember which is which though. But anyway, it might be it might be the opposite. It might be A. And B. Um Can't really tell we have different solutions but I can't really tell where that minimum is somewhere around in here but I can't tell you know which which is which is 1 2.5 of which is to say that That probably looks more like two there. So I'm thinking that's that's probably be, let's switch this just hey, B. All right. So what I did is I saw this for L substituted it back into here. And so I get see in terms of A. And B. And so I guess this is an F. Bar, it's C. Bar and I can pop that and I did that and I plotted it and we can see again, we have a minimum here. And if we look, it's when um A. Is five halves and B. Is too, And we can see that our minimum is gonna occur somewhere around 60. So it all looks consistent. So I'm pretty confident that that's correct.

So for this problem we've got this figure out the square base you have X. And X here and the height. Got the constraint said to you he X squared times H. Is equal to 12 solve in terms of X. We have H. Equals 12 X squared. Okay put over a squared we've got the objective equation here being seized equal to now. They did tell us that the only the top would be made of a different material. So we've got X squared times $2 for that. But everything else will be worth $1. So you've got $1 for this plus four X. H. Times to $1 for that. Simplifying this fully we've got three X squared plus four X. H. I'm going to substitute in for age so it's going to be three X squared plus four X times 12 or X squared simplifying that. We got three x squared Plus 48 Hover X. It's equal to see, right so now let's go ahead and take the derivative. So you've got C. Primes equal to six X By this. 48 over X squared. Okay, so from here, Set that equal to zero and That means we got six x equals 48 over ax squared, Multiplied by exportable, says we get six x cubed equals 48. Find out about six. We get x cubed equals eight. Therefore X must equal to when we keep route both sides. So that's the first thing. So it's going to be x equals two feet there. And then from here we've got that Y is equal to 12 over X squared, which means it's 12 over four chuckles three ft. So that script why that's supposed to be H. So the height is three ft Size of the squares, two ft and we're done.

19 talks about a rectangular container so the container will look something of the sort. Wow, let's, uh, market markets dimensions. So it's X, y and Z. The volume of the container will be X y Z, and the volume has already given us 3 20. So here we have X y Z s 3. 20. So here the value off value of Z would be 3 20 over ex wife. Let's keep this. Let's keep it. Keep it aside. We're going to use it in the eventual calculations. Now we're supposed thio. The cost for the base has given the cost to make the basis $5 for food area. Off the base will be X times y. So it's cost. Let's say C one x and Y will be. Since that is $5 per square feature, it will be five expire on. We have the area off the sides as we have two sides venture from the richer towards the left and right as Y and Z dimensions to the area will be two y z plus. We have the front and back, which is two x z, so the total area would be two times y Z plus X z Z can be taken out, so it is to see times five or six. So the cost C two off the sides will be its $4 per square feet, so that will be four times this value. So it will be a Z Viper sex on Let's let's Reza substitute the value of Z as 3 20 over X plus y. So that is eight times 3 20 over to your new were ex wife. So it's x Y over here on board were here we have five plus x, so we have 3 20 times eight. So that is equal to 560 and the open of the brackets. We have one over X plus one or why this is C two X, so the total cost would be see even proceed to illustrate the cost function, which is five expire plus 2 +5601 of our X plus one of our while. Let's find C X now to maximize this so we have five I plus 2 +560 minus one over extra square and let's find see why that will be five X minus +2560 over. Why Square? Let's equate both these +20 So we have five y is equal to 2560 over extra square on. We have five x as 2560 over. Why square? So from here we get the value of y as this is 256 0/5 X square. Let's put this value off. Why? Over here? So we have five x as 2560 ver y square. But just 256 0/5 X square, whole square. So this comes out as five x is equal to if we open up the denominator than this stone will come up. So we have 25 x rays to the powerful since that is a square and we have 2560 square. So run off the stone will cancel our we have five xs 25 x rays to the poor, 4/2 560 So from here we have five times five is 25 let's cross multiply. So we have 2560 Xs five x rays to the par four mhm dividing both sides by five. We get 512 x as X rays to the poor. Four. This means that X rays to the poor. Four minus 512 x zero Let's take X out, X Q minus 512 0. This means that either x zero, which is just not possible because the land cannot be zero and X Cube is 412. Uh, it means that the cube root off 512 is actually eight. So exquisite fixes eight. We gotta find corresponding why which is over here. So why would be to 56 year over five times eight square. So 256 0/5 years. 512 and 512 divided by 60 fours again. Eight. Okay, so X is it? And why is it? And there is no point off calculating whether there is a maximum minimum because there is only critical point and the question asks about having a minimum, so the dimensions of the container would be access it. Why is it on the Z would be from here? Uh, from this equation, the Z would be 3. 20/8 times eight so 3. 20/64 as nothing but five, So the value of Z would be five. So these other dimensions for the container toe have the minimum cost.

For this problem we are told that the cost of producing a rectangular boxes as follows. The sides cost $2 per square foot. The top $1 per square foot in the base. $4 per square foot were then told that if the volume of the boxes to be 10 cubic feet were asked to determine the dimensions that minimize the cost to begin. We'll have that. The cost will equal. Now we have the sides will be um One it is $2 per square foot. For the sides. We have $2 times. Now we have each side will be You have two times X. Y. for the front and back. You'll have two times Y. Z. For the other two sides. So we have two times two X. Y. Plus too easy for the sides. Then we have plus one times the cost of the top. The top is going to be X. Times E. And also we have that the base is $4 per square foot. So that will add on another four exit or plus five exit. So now we have our costs, we can multiply that too in so it's four X. Y. Plus four Y. Zed Plus five X. Set. And we're told that the volume must be 10 cubic feet. The volume is going to be X. Times Y. Times Z. And it must equal 10. So what we have to do here is figure out a way of minimizing the cost subject to the constraint X. Y. Z equals 10. So what we can do here is use the method of lagrange multipliers. So the first step is that we want to set the gradient of c. Radiant of C equal to lambda times the gradient of V. So we'll have first, when we take the partial derivative with respect to X, we'll have four Y plus five Z equals lambda times wise. It Then we have with respect to why we have four x plus four said equals lambda, X. Said. We then have uh with respect to Z. We have four Y plus five X equals lambda X. Y. And we also have our constraint X. Y. Z. equals 10. So we have a system of four equations with four unknowns one second here. So the solution that we get from that system of equations will be lambda equals for x equals two. Y equals five over to n. Z equals two.


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