91Ó°ÊÓ

Constants: - \(20\) cubic feet (Volume of the box) - \(0.10\) per square foot (Cost of the material for the sides) - \(0.25\) per square foot (Cost of the material for the top and bottom) Variables: - Length, width, and height of the box. We can use \(l\), \(w\), and \(h\) to represent them, respectively. Restrictions: All dimensions (\(l\), \(w\), and \(h\)) must be positive for a meaningful box.

We can calculate the cost of each side of the box and then add them up to get the total cost. Cost of the sides (2 lw sides and 2 wh sides): \(2lw(0.10)+2wh(0.10) = 0.20lw + 0.20wh\) Cost of the top and bottom (2 lh sides): \(2lh(0.25) = 0.50lh\) The total cost \(C\) is the sum of the cost of all sides: \(C = 0.20lw + 0.20wh + 0.50lh\)

Since the volume of the box is fixed, we have: \(V = lwh = 20\) We can solve for one of the variables, for example, \(h\): \(h = \frac{20}{lw}\) Now, we can substitute this into the cost function to eliminate \(h\) and express \(C\) as a function of the two independent variables \(l\) and \(w\). \(C(l, w) = 0.20lw + 0.20w\frac{20}{lw} + 0.50l\frac{20}{lw} = 0.20lw + 4w + 10l\)

To find the dimensions that minimize the cost, we need to find the critical points of the cost function \(C(l, w)\) and check if it is a minimum. Now, let's find the partial derivatives of \(C\) with respect to \(l\) and \(w\), and set them equal to zero: \(\frac{\partial C}{\partial l} = 0.20w - 10\frac{1}{w} = 0\) \(\frac{\partial C}{\partial w} = 0.20l - 4\frac{1}{l} = 0\) Solving the system of equations, we get: \(l = \sqrt[3]{50}\) \(w = \sqrt[3]{40}\) To verify that this critical point corresponds to a minimum, we can use the second derivative test. We need to find the second order partial derivatives and compute the Hessian determinant: \(\frac{\partial^2 C}{\partial l^2} = 10\frac{1}{w^2} > 0\) \(\frac{\partial^2 C}{\partial w^2} = 4\frac{1}{l^2} > 0\) \(\frac{\partial^2 C}{\partial l \partial w} = 0\) Hessian determinant: \(D =\frac{\partial^2 C}{\partial l^2}\frac{\partial^2 C}{\partial w^2} -(\frac{\partial^2 C}{\partial l \partial w})^2 = 40\frac{1}{l^2w^2} > 0\) Since the Hessian determinant is positive and the second partial derivative with respect to \(l\) is also positive, this critical point corresponds to a minimum. Therefore, the dimensions that minimize the cost of a box are: Length: \(l = \sqrt[3]{50}\) Width: \(w = \sqrt[3]{40}\) Height: \(h = \frac{20}{lw} = \frac{20}{(\sqrt[3]{50})(\sqrt[3]{40})} = \sqrt[3]{\frac{1}{100}}\) These dimensions will result in the minimum cost for the manufacturer.