Q. 15

Constructing Rain Gutters A rain gutter is to be made of aluminum sheets that are 12 inches wide by turning up the edges 90. See the illustration.

(a) What depth will provide maximum cross-sectional area and hence allow the most water to flow?

(b) What depths will allow at least 16 square inches of water to flow?

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Answer

(a) The depth that will provide maximum cross sectional area is $3 i n c h e s$

(b) The depth that allow up to $16 i n^{2}$ to flow are depth of $4 i n c h e s a n d 2 i n c h e s$

1Step 1. Given information

The width of the sheets for the rain gutters $12 i n c h e s$

The angle at the edges of the fformed rain gutter $= 90 °$

2Step 2. (a) What depth will provide maximum cross-sectional area and hence allow the most water to flow?

Derive a function for the cross sectional area and find the values of the variable of the function at the maximun point by differentiation

Let y represent the depth of the rain gutter that provide maximum cross section,and let x represents the width of the rain gutter we have

$x + 2 y = 12 x = 12 - 2 y$

The cross sectional area $A = d e p t h x w i d t h$

$A = x \times y = (12 - 2 y) \times y = 12 y - 2 y^{2}$

At maximum cross sectional area, $A_{m a x}$ ,we have

$\frac{d A}{d y} = 0 \frac{d A}{d y} = \frac{d (12 y - 2 y^{2})}{d y} = 12 - 4 y = 0$

Therefore,at $A_{m a x}$ , $12 - 4 y = 0$

$4 y = 12 y = \frac{12}{4} = 3$

Therefore,The depth that will provide maximum cross sectional area is $= 3 i n c h e s$

3Step 3. (b) What depths will allow at least 16 square inches of water to flow?

The depth that will be allow $16 i n^{2}$ of water to flow is given as follows

$A = x \times y x \times y = 16 x = 12 - 2 y A = (12 - 2 y) \times y 12 y - 2 y^{2} = 16$

dividing both sides by 2 gives;

$\frac{(12 y - 2 y^{2})}{2} = \frac{16}{2} 6 y - y^{2} = 8 y^{2} - 6 y + 8 = 0 (y - 4) (y - 2) = 0 y = 4 o r y = 2$

Therefore,The depth that allow up to $16 i n^{2}$ to flow are depth of $4 i n c h e s o r 2 i n c h e s$