a. squares with sides of length x are cut out of each corner of a rectangular piece of cardboard measuring…

Explanation:

Step1: Set up volume formula for part b

Let the side - length of the square cardboard be (s) and the side - length of the square cut from each corner be (x). The length and width of the base of the box will be (s - 2x), and the height of the box is (x). So the volume (V(x)=(s - 2x)(s - 2x)x=x(s - 2x)^{2}=x(s^{2}-4sx + 4x^{2})=s^{2}x-4sx^{2}+4x^{3}).

Step2: Find the derivative of (V(x))

Using the power rule ((x^n)^\prime=nx^{n - 1}), we have (V^\prime(x)=s^{2}-8sx + 12x^{2}).

Step3: Set the derivative equal to zero

To find the critical points, we set (V^\prime(x)=0), so (12x^{2}-8sx + s^{2}=0). Using the quadratic formula (x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}) for the quadratic equation (ax^{2}+bx + c = 0) (here (a = 12), (b=-8s), (c = s^{2})), we get (x=\frac{8s\pm\sqrt{64s^{2}-48s^{2}}}{24}=\frac{8s\pm\sqrt{16s^{2}}}{24}=\frac{8s\pm4s}{24}). The solutions are (x_1=\frac{8s + 4s}{24}=\frac{s}{2}) and (x_2=\frac{8s-4s}{24}=\frac{s}{6}). But (x=\frac{s}{2}) will make (s - 2x = 0), so we discard it. We use (x=\frac{s}{6}).

Step4: Find the maximum volume

Substitute (x = \frac{s}{6}) into the volume formula (V(x)). (V(\frac{s}{6})=\frac{s}{6}(s-2\times\frac{s}{6})^{2}=\frac{s}{6}(\frac{2s}{3})^{2}=\frac{s}{6}\times\frac{4s^{2}}{9}=\frac{2s^{3}}{27}).

Answer:

(\frac{2s^{3}}{27})