Page 102 - How To Solve Word Problems In Calculus

2
2
h + 10 = 20 2
2
h + 100 = 400
2
h = 300
√ √
h = 300 = 10 3
10 − x y
Observing similar triangles, = √ . Multiplying, we
10 10 3
get
√
10y = 10 3(10 − x)
√
y = 3(10 − x)

The area as a function of x may be written

√
A(x) = 2x 3(10 − x)
√
2
= 2 3(10x − x )

From the diagram it is clear that 0 ≤ x ≤ 10

Step4
√
A (x) = 2 3(10 − 2x)

√
0 = 2 3(10 − 2x)
0 = 10 − 2x

2x = 10
x = 5

√ √
The corresponding value of y = 3(10 − x) = 5 3. Since the
dimensions of the rectangle were 2x and y, the largest possible
√ √
area is 10 × 5 3 = 50 3.

Step5 (optional)
If x = 0or x = 10, the area of the rectangle is 0. Therefore,
the absolute maximum area occurs at x = 5.

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