Optimization word problem - minimizing surface area to find least expensive tank

In summary, the question asks for the cost of the least expensive tank with a rectangular base and rectangular sides, given a width of 4 meters and a volume of 36 cubic meters. To find this cost, we need to minimize the surface area of the tank, which can be expressed as SA = 8H + 8LH + 16L, where H is the height and L is the length of the tank. Using the given information, we can simplify this equation to H + 4L = 9. By setting the derivative of this equation equal to 0, we can solve for the optimal values of H and L, which we can then use to calculate the final cost of the tank.
  • #1
cahsuhdee
3
0

Homework Statement


A tank with a rectangular base and rectangular sides is to be open at the top. It is to be constructed so that its width is 4 meters and its volume is 36 cubic meters. If building the tank covers $10 per square meter for the base and $5 per square meter for the sides, what is the cost of the least expensive tank?


Homework Equations


Surface Area of a Rectangular Prism 2(LW) + 2(LW) of other set of sides + LW of base
Volume = LWH



The Attempt at a Solution


First thing.. come up with the primary Equation... and since I'm trying to minimize Surface Area... SA = 8H (smaller sides) + 8LH (larger sides) + 16L (base)

Secondary Equation, in order to manipulate the variables so I can plug them into the primary equation... Volume = LWH = 36

And then... after plugging whatever expression I get for the variable into the primary eq... I know I do the derivative.. set it equal to 0.. then solve.. or at least that's what we HAVE been doing for previous problems.. But before it was always something like.. "Find the dimensions of blah which would maximize the volume" .. and they never had this many variables.. so really I'm not sure what I'm doing, after the mere realization that somehow I use the volume to apply it to the surface area, then multiply it by cost to find how much it'd be. I'd be thankful for any sort of help!
 
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  • #2
cahsuhdee said:
I'm trying to minimize Surface Area

First thing you need is to be clear on what you are to minimize. It isn't surface area. Read the problem statement again.

Write an equation for what it is you are trying to minimize and use the given information to simplify the equation to a single variable. Once you have that, set the derivative equal to zero and solve for the value of the variable.
 

FAQ: Optimization word problem - minimizing surface area to find least expensive tank

1. What is an optimization word problem?

An optimization word problem involves finding the maximum or minimum value of a specific quantity based on given constraints. In this case, we are trying to minimize the surface area of a tank while maintaining a certain volume.

2. Why is minimizing surface area important in finding the least expensive tank?

The surface area of a tank directly affects the cost of materials and construction. By minimizing the surface area, we can reduce the amount of materials needed and ultimately decrease the cost of the tank.

3. What are the steps involved in solving an optimization word problem?

The steps involve identifying the objective function (in this case, minimizing surface area), writing down the constraints (such as volume and shape requirements), finding and solving the derivative of the objective function, and checking for critical points and endpoints to determine the minimum value.

4. How does calculus play a role in solving this optimization word problem?

Calculus is used to find the derivative of the objective function, which helps us determine the minimum value. The concept of critical points and endpoints is also essential in solving optimization problems using calculus.

5. What are some real-life applications of optimization word problems?

Optimization word problems are commonly used in various fields such as engineering, economics, and physics. Examples include finding the most cost-effective design for a bridge, maximizing profits for a company, and minimizing energy consumption in a building.

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