Find the maximum value of box?

a square sheet of metal has sides of length 8cm. each square piece is removed from each corner and the remaining piece is bent into the form of an ope

a square sheet of metal has sides of length 8cm. each square piece is removed from each corner and the remaining piece is bent into the form of an open box. find the maximum value of box.

or:a square sheet of metal has sides of length 8cm. each square piece is removed from each corner and the remaining piece is bent into the form of an open box. find the maximum value of box.

or:What's to be maximised? The volume of a cuboid with a square base and a height.the base material is given with a square sheet of metal that has sides of length 8cm.The relation between the side length of the base b and the height h is obviously:b = 8 - 2hSo, the cuboid will have the volume v = h * (8 - 2h)^2.Our target function is then:f (x) = x (8 - 2x)^2The problem-related domain of the function is determined by the set of sensible values for x which we feed into it: A height of h = 4. Thus we are only interested in the values in the closed interval [0, 4].We can write this:D = { x | 0 < x < 4 }.We want to know, where this function has its maximum value in the problem-related domain. We, therefore have to find the roots of the 1st derivative:f '(x) = 2x (8 - 2x) * -2 + (8 - 2x)^2f '(x) = -4x (8 - 2x) + (8 - 2x)^2f '(x) = (8 - 2x) ( -4x + 8 - 2x)f '(x) = (8 - 2x) ( -6x + 8)The 1st derivative is zero, where f (x) has its extrema. So,(8 - 2x) ( -6x + 8) = 0 ;; divide by 2(4 - x) ( -3x + 4) = 0gives us at a glance:x1 = 4, x2 = 4/3.x1 goes away (as this would leave us with 4 little square pieces of 4cm side length each, but no base for the box, which is why we excluded it from the domain in the first place).Is there at f (x2) a maximum or a minimum? The second derivative will tell us:f ''(x) = -3 (4 - x) + (-1)( -3x + 4)f ''(x) = 6x - 16and f ''(4/3) = -8 < 0 --> Maximum!Thus the side length of the square pieces that have to be removed from each corner ish = 4/3 cm = 1.33 cm, the side length of the base isb = (8 - 2.66) cm = 5.34 cm, and, finally, the volumev = 1.33 * 5.34^2 = 37.93 cm^3.

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