450
Chapter 4 | Applications of Derivatives
that approaches infinity at the ends, it must have an absolute minimum at some x ∈ (0, ∞). This minimum must occur at a critical point of S . The derivative is S ′( x ) = − 864 x 2 +2 x . Therefore, S ′( x ) =0 when 2 x = 864 x 2 . Solving this equation for x , we obtain x 3 =432, so x = 432 3 =6 2 3 . Since this is the only critical point of S , the absolute minimum must occur at x =6 2 3 (see Figure 4.70 ). When x =6 2 3 , y = 216 ⎛ ⎝ 6 2 3 ⎞ ⎠ 2 =3 2 3 in. Therefore, the dimensions of the box should be
x =6 2 3 in.
3 in.
and y =3 2
With these dimensions, the surface area is S ⎛ ⎝ 6 2 3 ⎞ ⎠ = 864 6 2 3 + ⎛ ⎝ 6 2 3 ⎞ ⎠ 2 =108 4 3 in. 2
Figure 4.70 We can use a graph to determine the dimensions of a box of given the volume and the minimum surface area.
4.36 Consider the same open-top box, which is to have volume 216 in. 3 . Suppose the cost of the material for the base is 20 ¢ /in. 2 and the cost of the material for the sides is 30 ¢ /in. 2 and we are trying to minimize the cost of this box. Write the cost as a function of the side lengths of the base. (Let x be the side length of the base and y be the height of the box.)
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