Section 14.7: Max Min Problems

SOLUTION Let the length, width, and height of the box (in meters) be \(x, y\), and \(z\). Then the volume of the box is \[ V = xyz \] We can express \(V\) as a function of just two variables \(x\) and \(y\) by using the fact that the area of the four sides and the bottom of the box is \[ 2xz + 2yz + xy = 12 \] Solving this equation for \(z\), we get \(z = (12 - xy)/[2(x + y)]\), so the expression for \(V\) becomes \[ V = xy \frac{12 - xy}{2(x + y)} = \frac{12xy - x^2y^2}{2(x + y)} \] We compute the partial derivatives: \[ \frac{\partial V}{\partial x} = \frac{y^2(12 - 2xy - x^2)}{2(x + y)^2} \quad \frac{\partial V}{\partial y} = \frac{x^2(12 - 2xy - y^2)}{2(x + y)^2} \] If \(V\) is a maximum, then \(\partial V/\partial x = \partial V/\partial y = 0\), but \(x = 0\) or \(y = 0\) gives \(V = 0\), so we must solve the equations \[ 12 - 2xy - x^2 = 0 \quad 12 - 2xy - y^2 = 0 \] These imply that \(x^2 = y^2\) and so \(x = y\). (Note that \(x\) and \(y\) must both be positive in this problem.) If we put \(x = y\) in either equation we get \(12 - 3x^2 = 0\), which gives \(x = 2\), \(y = 2\), and \(z = (12 - 2 \cdot 2)/[2(2 + 2)] = 1\). We could use the Second Derivatives Test to show that this gives a local maximum of \(V\), or we could simply argue from the physical nature of this problem that there must be an absolute maximum volume, which has to occur at a critical point of \(V\), so it must occur when \(x = 2, y = 2, z = 1\). Then \(V = 2 \cdot 2 \cdot 1 = 4\), so the maximum volume of the box is 4 m\(^3\).