Solving a Costly Word Problem: Finding the Cheapest Container

[nspskier]

I'm having problems setting up this problem (word problems are my worst downfall!)

a rectangular container with an open top is to have a volume of 10 meters cubed. The length of its base is 2x the width.
The cost of the base is $12 per sq meter. The cost of the sides is $5 per sq meter. find the cost fo the materials for the cheapest container.

 

[NateTG]

Can you write equations for the volume and cost of the container in terms of it's dimensions? Can you write equatiosn for the dimensions in terms of each other?

 

[M98Ranger]

Word problems can definitely be tricky, but with some practice and a clear approach, they can become easier to solve. In this particular problem, we are given the volume and dimensions of a container, as well as the cost of materials for the base and sides. Our goal is to find the cost of materials for the cheapest container.

To set up this problem, we need to start by defining our variables. Let's let x represent the width of the base. Since the length is 2x, we can represent the length as 2x as well.

Next, we can use the formula for the volume of a rectangular prism to set up an equation: Volume = length x width x height. In this case, our volume is given as 10 meters cubed, so we can write the equation as 10 = (2x)(x)(h), where h represents the height of the container.

Now, we need to use the given information about the cost of the base and sides to set up an equation for the total cost. The cost of the base is $12 per sq meter, so the cost of the base will be 12 times the area of the base, which is (2x)(x). Similarly, the cost of the sides is $5 per sq meter, so the cost of the sides will be 5 times the surface area of the sides, which is 2(2xh) + 2(xh). This gives us the equation: Cost = 12(2x^2) + 5(4xh + xh).

Now, we can use the equation for volume to solve for h, and then substitute that value into the equation for cost. This will give us a function for the total cost in terms of x: Cost = 24x^2 + 20xh.

To find the cheapest container, we want to minimize the cost function. We can do this by finding the critical points of the function, which are the points where the derivative is equal to 0. After taking the derivative and setting it equal to 0, we can solve for x and h.

Once we have the values for x and h, we can substitute them back into the cost function to find the minimum cost. This will give us the cost for the cheapest container.

In summary, to solve this word problem, we need to define our variables, set up equations for volume and cost, and then use calculus