A rectangular storage container without a lid is to have a volume of 10 m3. The length of its base is twice the width. Material for the base costs $15 per square meter. Material for the sides costs $9 per square meter. Find the cost (in dollars) of materials for the least expensive such container. (Round your answer to the nearest cent.)$

To solve this problem, we need to use calculus to minimize the cost function.

Step 1: Define the variables Let's denote the width of the base as x (in meters), the length as 2x (since it's twice the width), and the height as h (in meters).

Step 2: Express the volume in terms of x and h The volume V of the rectangular container is given by the formula V = length * width * height. Substituting our variables, we get 10 = 2x * x * h, which simplifies to 10 = 2x^2h.

Step 3: Express h in terms of x From the volume equation, we can express h in terms of x: h = 10 / (2x^2) = 5 / x^2.

Step 4: Define the cost function The cost C of the materials is the sum of the cost of the base and the cost of the sides. The base has area 2x^2 and costs $15 per square meter, so its cost is 15 * 2x^2 = 30x^2. Each side has area xh or 2xh and costs$9 per square meter, so the total cost of the sides is 9 * 2xh + 9 * 2xh = 36xh. Substituting h = 5 / x^2, we get C = 30x^2 + 36x * 5 / x^2 = 30x^2 + 180 / x.

Step 5: Minimize the cost function To find the minimum cost, we take the derivative of C with respect to x, set it equal to zero, and solve for x. The derivative is C' = 60x - 180 / x^2. Setting this equal to zero gives 60x = 180 / x^2, or x^3 = 3. The cube root of 3 is approximately 1.44, so x = 1.44 meters.

Step 6: Find the minimum cost Substituting x = 1.44 into the cost function gives C = 30 * (1.44)^2 + 180 / 1.44 = 62.37 + 125 = $187.37. So the least expensive such container costs approximately$187.37.