Page 64 - Calc for the Clueless

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As before, the volume = (1)(w)(h). With a square base, V = x y. A box with no top has five surfaces—a square
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bottom and four sides, all of which have the same dimensions. The surface area S = x + 4xy.
Let's do A.

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The box is 2 by 2 by 1 feet. S = x + 4xy = (2) + 4(2)(1) = 12 square feet. S"= 2 + 32x . S"(2)= 2 + 32/8 = 6,
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which is positive, indicating a minimum.
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Now let's do B. In this case we are maximizing V = x y if the surface area S = x + 4xy = 12. y = (12 - x )/4x.
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The box is 2 by 2 by 1 feet, and its maximum volume is 4 cubic feet. Of course, we knew this from the first part
of the problem. To check this, V" = -3x/2. V"(2) = -3, which is negative, indicating a maximum.

Let's try one that really doesn't have a picture.

Example 5—

An orchard has 50 apple trees per acre. The average number of apples per tree is 990. For every additional tree
per acre planted, each apple tree will give 15 less apples. Find the number of additional trees per acre that
should be planted to give the largest number of apples per acre.

The total number of apples equals the product of the number of trees and the number of apples per tree. If we
add x trees, we must subtract 15x apples per tree. There are x + 50 trees and 990 - 15x apples per tree. The total
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apples A = (x + 50)(990 - 15x) = 49,500 + 240x - 15x . A' = 240 - 30x = 0. x = 8 trees.
Note