How big should the square be cut to the sandbox able to hold the most sand?
Suppose you have an 8-foot by 12 feet (8 'x 12'), copper foil. If you were to cut congruent squares from each corner, you can fold the four sides. If you then soldier from the cut edges together you would have a container that can be used as a litter box. How big should the square cut to make the litter box able to contain the most sand? Be sure to show how you got your answer
This is commonly called an optimization problem. You want to maximize the volume of the resulting litter box, which is a function V (x) x length of the sides of the square cut (This is known because this is the only variable in the problem). First draw a diagram and deduce the formula V (x) = x (8 – 2x) (12 – 2x). From your diagram, you should see that x is the height of the resulting litter box, and decreases the length and width of the litter box for x in each corner, making a total decrease of 2 times per dimension. Then just multiply the x ^ 3 length by width by height to obtain V (x) as above. The function V (x) is a cubic V (x) = 4x ^ 3 – 40x ^ 2 + 96x. Use the first derivative to find critical points: V '(x) = 12x ^ 2 – 80x + 96. The critical points occur where the first derivative is 0 (the tangent to the maximum of a smooth curve is horizontal, so that the derivative is not 0). Divide both sides by 12 to get x ^ 2 – (20 / 3) x + 8 = 0. Because the fraction, it is easier to solve this equation with the quadratic formula, x = ((20 / 3) + / – Sqrt [400 / 9 - 32]) / 2 = 10.3 + / – 2Sqrt [7]. Both values are positive, but sanity check tells us that 10 / 3 + 2Sqrt [7] is higher than 8, so this value is not physical. Then, just check the 10 / 3 – 2Sqrt [7] for if it is up to the volume function. If a point on a smooth curve is a maximum, then the slopes of tangent lines to the graph around that point goes from positive to negative by inspection, so that the resulting decrease around this point. Therefore, the derivative of the derivative (second derivative) must be negative in that point. Length test our first find the second derivative: V''(x) = 24x – 80. We connect our value for x, and obtain V''(10 / 3 – 2Sqrt [7]) = 8 * (10 / 3 – 2Sqrt [7]) – 80 = -160 / 3 – 16Sqrt [7], which is negative. Therefore, this length is a maximum for the function of volume.
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