A very High Level Math Problem!!!! Please help me out!!!?

An open-top box is to be made so that its width is 4 ft and its volume is 40 ft^3. The base of the box costs $4/ft^2 and the sides cost $2/ft^2.





a) Express the cost of the box as a function of its length (l) and height (h)


b) Find a relationship between l and h.


c) Express the cost as a function of h only


d) Give the domain of the cost function


e)use a graphing claculator or computer to approximate the dimensions of the box having least cost.





If you do this and show your work i will love you!!!!! Please i really need some help. Its due 2moro and i did it but i think it is wrong!!!!!

A very High Level Math Problem!!!! Please help me out!!!?
Hi,





a) Express the cost of the box as a function of its length (l) and height (h)





Cost = $4/ft²(LW) + $2/ft²((2L + 2W)H)


Cost = $4/ft²(L*4) + $2/ft²((2L + 2*4)H)


Cost = 16L + 2(2HL + 8H)


Cost = 16L + 4HL + 16H





b) Find a relationship between l and h.


Volume = LWH


40 = L(4)H


10 = LH





10


---- = H


..L





c) Express the cost as a function of h only





Cost = 16L + 4HL + 16H


Cost = 16L + 4(10/L)L + 16(10/L)


Cost = 16L + 40 + 160/L





d) Give the domain of the cost function


L %26gt; 0





e)use a graphing calculator or computer to approximate the dimensions of the box having least cost.





Y = 16x + 40 + 160/x





If this is graphed, it has a minimum at (3.162,141.19). this means the minimum cost is when the length is 3.162 feet and the cost of the box is $141.19. The height is 10/L = 10/3.162 = 3.162.





I hope that helps!! :-)
Reply:Okay, what do we know. We know that the width is 4 and the volume is 40, so what do we have left? We want cost right? Well, cost is going to be in terms of square feet, so our answer has to be in square feet. Remember that, if your units end up being wrong, you know you're not doing it right. So, if you know the width, you are going to need some kinda formula right? Don't worry about the math yet, just do this:





How much will the box cost? The total cost is the cost of the base plus the cost of each side.





Now there's something that we can put in a formula!





Total cost = base cost + sides cost





Now, let's just look at the base for now. We know the width is 4, and the length is L, and the base costs $4 per square foot. So...





Base cost = $4 * 4ft * L





Now, there are two sets of "sides" let's say the front and back, and then the other two are "sides". So we'll some equation like this for the front and back. (Remember 4 wide, $2 per sq ft)





Front and Back Cost = 4ft (width) * H * 2 for each side * $2


Two Sides Cost = L * H * 2 (for each side) * $2





Total Sides cost = (4ft * H * 2 * $2) + (L * H * 2 * $2)





And now we add that to the base cost to get total cost as a function of H and L





f(H, L) = ($4 * 4ft * L) + ((4ft * H * 2 * $2) + (L * H * 2 * $2))


f(H,L) = (16L) + ((16H) + (4LH))


f(H,L) = 16L + 16 H + 4LH


(I think this is your "a" question.)





But wait! It's gotta be 40 cubic feet right?!?! Don't worry. So we need some kind of way to make sure it stays that way. We need H to change with L so that it's always 40 cubic feet. So:





40 cubic feet = 4ft * H * L -%26gt; or


10 square feet = H * L or


H = 10 square feet / L and


L = 10 square feet / H





So, there's a relationship between the two (aka, question 'b')





Now, just take our original f(H,L) and drop in our relationship for L= 10/H for each L we see:





f(H,L) = 16L + 16 H + 4LH


f(H) = 16(10/H) + 16H + 4(10/H)(H)


f(H) = 160/H + 16H + 40





Now, we have a formula only in terms of H (aka 'c')





Now, we can't have a box with a height of 0, but we could make this box as tall as we want, as long as H = 10/L. It's just that as the box gets taller, L gets smaller. So really, the possible values for the height of the box (aka the domain of the equation, and the answer to 'd') is anything greater than 0.





As for the last one, just graph it, I can't find my graphing calculator, and that's the easy part. I hope that I explained this well enough for you to follow.