Page 34 - How To Solve Word Problems In Calculus

Solutions to Supplementary Problems

1. Let x = the smaller number and y = the larger number. Their
product is P = xy. Since y − x = 15, y = x + 15. By substitution,

P (x) = x(x + 15)
2
P (x) = x + 15x
2. y

x

200
P = 2x + 2y. Since A = xy = 200, y = . By substitution,
x

200
P = 2x + 2
x
400
P (x) = 2x + x > 0
x
3. y

x x x x

y

Let L = length of fence used. L = 4x + 2y. The enclosed area,
1000
xy = 1000, so y = . It follows that
x

1000
L = 4x + 2
x
2000
L (x) = 4x + x > 0
x
4. Let (x, y) represent the point on the circle corresponding to the
upper right corner of the rectangle. The length of the rectangle will
then be 2x and the height y.

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