A rectangular storage container with an open top is to have a volume of $ 10 m^3 $. The length of its base is twice the width. Material for the base costs \$ 10 per square meter. Material for the sides costs \$ 6 per square meter. Find the costs of materials for the cheapest such container.
min cost $=\$ 163.54$
suppose you want to find the total minimum cost of materials for a storage container Given the volume of this container is 10 m3, the length is twice the width. The cost for based material is 10 per square meter And the cost for side material is six per square meter. First we want to define our variables let w be the with L. B. The length of the base and hb the height of the rectangular container. Next we identify the objective function. Now this is the function which is minimized or maximized. So in this case we want to get the total minimum cost. So the objective function will be the total cost C. Which is just the sum for the cost for based material and cost for side material. Note that the surface area of a rectangular prison is equal to the area of the base plus the area of the two sides plus the area of the front and the back. And so based on this figure, we have the surface area equal to all times W plus twice HL plus twice. HDW given the cost for the base inside material, we have the total cost Which is equal to 10 times L. W. That's six times the sum of 2. 2 HW. And so our objective function is 10 times L. W plus six times The sum of two HLN 2 HW. Now we want to write this objective function in terms of one variable only. We will use the fact that the length is twice its width and so we have L which is equal to two W. And we know that the this is length times width times height, this is equal to 10. So if L is to W then we have the equal to two W Times W times age, this is equal to 10. And so we have two w squared times ages 10. and solving for eight, you get H equal to five over w squared, you seeing Ellie close to W and H equals five over W squared. We have See which is equal to 10 times two W Times W. Plus We have six times two times 8 reaches five over W sq times L which is to W plus two times a judge's five over W squared times W. And so simplifying this, we have see which is equal to 20 W squared plus six times you have 20 over W plus 10 over W. and simplifying further, we get c equal to 20 W squared plus six times 30 over W which will give us C equal to 20 W squared Plus 180 Times W raised a -1. And so this is our objective function in terms of one variable from here, we want to find the critical point of C to do this, you first find, see prime, that's equal to 40 W minus 180 W race too negative too. That's the same as 40 W -1, 80 over W squared. Next we want to set C Prime 20 and solve for W. So we have 40 W -180 over w squared. This is equal to zero Solving for W. We have 40 W. This is equal to 1 80 over W squared. And so we get W to the third power equal to 180 over 40 or you have nine over to And so W is just The Cube rid of nine over to which is our critical point. And to know if this critical point minimizes, see we apply second derivative test. So if C prime is 40 W minus 1 80 times W raised negative too. Then we have C double prime. This is equal to 40 plus. That's 3 60 W Raise 2 -3. That's the same as 40 plus 360 over W cube. So since W is are the critical point is cube rid of Of nine over to. Then evaluating see double prime. You using this W You have see W of the Cuban of nine over to this is just 40 plus. 360 over the cube of the Q bridge. Ah nine over to which is just 40 plus 3 60 times to overnight just equal to 1 20. And since this is greater than zero, the value of the second derivative then you see that my second derivative test, C is minimum when W is the cube rid of nine over to. Therefore the total minimum cost is this is C. Of The Cube root of nine over to. This is equal to 20 times the Square of the Cube rid of nine over to Plus 1 80 over the Q. Birth of 9/2, and this is approximately $163 and 54 cents.