If you want to produce an open-top box from a 16 in by 22 in flat piece of cardboard, what is the maximum possible volume (in in3) of the box? Round your answer to the nearest whole number and omit units.

Answer:480Step-by-step explanation:Since, for making a box from a cardboard,We need to cut four congruent pieces from each corner of the cardboard,Let x be the side of a piece ( in inches ),Given,The dimensions of the cardboard are 16 in by 22,So, the dimension of the box would be (16-2x) in by (22-2x) in by x in,Thus, the volume of the box,[tex]V(x)=(16-2x)(22-2x)x=4x^3-76x^2+352x[/tex]Differentiating with respect to x,[tex]V'(x) = 12x^2-152x+352[/tex]Again differentiating with respect to x,[tex]V''(x) = 24x-152[/tex]For maxima or minima,V'(x) = 0[tex]\implies 12x^2-152x+352=0[/tex]By the quadratic formula,[tex]x=\frac{-(-152)\pm \sqrt{-152^2-4\times 12\times 352}}{24}[/tex][tex]x=\frac{152\pm \sqrt{6208}}{24}[/tex][tex]\implies x\approx 9.62 \text{ or }x\approx 3.05[/tex]Since, for x = 9.62, V''(x) = positive,While for x = 3.05, v''(x) = negative,Hence, volume is maximum for x = 3.05,And, maximum volume, [tex]V(3.05) = 4(3.05)^3-76(3.05)^2+352(3.05)=480.1005\approx 480\text{ cube in}[/tex]