Abox (with no top) is to be constructed from a piece of cardboard of sides a and b by cutting out squares of length h from the corners and folding up the sides. find the value of h that maximizes the volume of the box if a = 11 and b = 8.

h = (19 - sqrt(97))/6, which is approximately 1.525190366 The volume of the box will be V = lwh And l will be a - 2h And w will be b - 2h So using the above, the volume of the box will be V = lwh V = (a - 2h)(b - 2h)h V = (11 - 2h)(8 - 2h)h V = (88 - 22h -16h + 4h^2)h V = (88 - 38h + 4h^2)h V = 88h - 38h^2 + 4h^3 Since you're looking for a maximum, that screams "First derivative" So let's calculate the first derivative of the function and solve for 0. V = 88h^1 - 38h^2 + 4h^3 V' = 1*88h^(1-1) - 2*38h^(2-1) + 3*4h^(3-1) V' = 1*88h^0 - 2*38h^1 + 3*4h^2 V' = 88 - 76h + 12h^2 We now have a quadratic equation. So using the quadratic formula with A=12, B=-76, and C=88, calculate the roots as: (19 +/- sqrt(97))/6 which is approximately 1.525190366 and 4.808142967 We can ignore the 4.808142967 value since although it does indicate a slope of 0, it produces a negative width and is actually a local minimum of the volume function. So the optimal value of h is (19 - sqrt(97))/6, which is approximately 1.525190366

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