Optimization, again: construct a rectangular storage container that has....

[whiteti]

I want to construct a rectangular storage container that has a square base, a closed top, and a volume of 9ft³. If the material for the base costs $5/ft² and the material for the sides and top costs $4/ft², what is the cost of the cheapest such container?

I'm stuck, how do I set up my equation?

 

[Deleted member 4993]

I want to construct a rectangular storage container that has a square base, a closed top, and a volume of 9ft³. If the material for the base costs $5/ft² and the material for the sides and top costs $4/ft², what is the cost of the cheapest such container?

I'm stuck, how do I set up my equation?

What do you need to "find"?

What are "given" to you?

What are the connections between the "givens" and the "find"?

 

[DrPhil]

I want to construct a rectangular storage container that has a square base, a closed top, and a volume of 9ft³. If the material for the base costs $5/ft² and the material for the sides and top costs $4/ft², what is the cost of the cheapest such container?

I'm stuck, how do I set up my equation?

First step is to assign names for the variables - such as
Let x = side of square base
.....h = height of container

Then make equations for the needed areas - to find the total cost, you will need the area of the base, and also the combined area of the sides and top.

Let us see how far you get.

 

[whiteti]

What do you need to "find"?

What are "given" to you?

What are the connections between the "givens" and the "find"?

Im at
V=xy
and

5y+4x+5y+4x = 10y+4x

solving for y is y=2/5x

so substituting back into V=xy
V=x(2/5x) = 2/5 x²

not sure where to go from here, or if this is even correct

 

[Deleted member 4993]

Im at
V=xy ................................................... What are 'x' and 'y'?
and

5y+4x+5y+4x = 10y+4x

solving for y is y=2/5x

so substituting back into V=xy
V=x(2/5x) = 2/5 x²

not sure where to go from here, or if this is even correct

.

 

[whiteti]

x is the base
y is the top and sides
V is being minimized?

 

[Deleted member 4993]

x is the base........................................does not make sense - x is what property of base - length of one side? cost? thickness? ... what?

y is the top and sides........................................ similarly does not make sense

V is being minimized? ..................................... what is V - related to box?

.

 

[srmichael]

x is the base
y is the top and sides
V is being minimized?

I think you meant, x is the length of the sides of the base (they are both x since the base is a square).

So V = lwh or x*x*y or x²y.

Next, you need the surface area formula using the cost of the materials. What is the surface area formula of your container? And no it's not 10y + 4x.

 

[whiteti]

.

good point, I wasn't really thinking

So i need the area of the base, sides and top separately?
How does the given value for volume come into play here?

I want x to be whatever the base has to be

and y to be relevant the top and sides

 

[whiteti]

ok, Ive taken a new approach

Cost of base: 5wl
Cost of front: 4lh
Cost of back: 4lh
Cost of left side: 4wh
Cost of right side: 4wh
Cost of top: 4wl

Cost= 5wl+2(4lh)+2(4wh)+4wl
= 5wl+8lh+8wh+4wl
=9wl+8lh+8wh

now I'm stuck again...

 

[DrPhil]

ok, Ive taken a new approach

Cost of base: 5wl
Cost of front: 4lh
Cost of back: 4lh
Cost of left side: 4wh
Cost of right side: 4wh
Cost of top: 4wl

Cost= 5wl+2(4lh)+2(4wh)+4wl
= 5wl+8lh+8wh+4wl
=9wl+8lh+8wh

now I'm stuck again...

Since the base is a square, w = l = x (feet). Thus (for instance),
......Cost of base = ($5/ft^2) * x^2
......Cost of 4 sides = ($4/ft^2) * perimeter * h
......Cost of top = . . .

There are two unknowns in the equation for total cost: x and h

How about the volume? You can use the fact that V = 9 ft^3 to find a relationship between x and h. Then you can convert the Cost equation to a single variable, and finally minimize that cost function. CARRY ON!

You may notice that I write units on EVERY number. That is a tool to be sure I do the correct arithmetic: area must be the product of two lengths, and Volume must be cubic feet. Cost has to turn out to be $.