Question

a rain gutter is made from sheets of aluminum that are 10 inches wide by turning up the edges to form right angles. Determine the depth of the gutter that will maximize its cross-sectional area and allow the greatest amount of water to flow.What is the maximum?

Answer 1

After turning the edges of the sheet, the dimensions will be x and 10 - 2x, so the cross-sectional area will be:

`\begin{gathered} A=x(10-2x) \\ A=10x-2x^2 \end{gathered}`

In order to maximize this area, let's find the x-coordinate of the vertex, using the formula:

`x_v=-(b)/(2a)`

Where a and b are coefficients of the quadratic function in the standard form:

`\begin{gathered} y=ax^2+bx+c \\ A=-2x^2+10x+0 \\ a=-2,b=10,c=0 \\ \\ x_v=-(b)/(2a)=-(10)/(-2\cdot2)=(10)/(4)=2.5 \end{gathered}`

So the depth of the gutter is 2.5 inches.

And the maximum cross-sectional area is:

`\begin{gathered} A=10x-2x^2 \\ A=10\cdot2.5-2\cdot2.5^2 \\ A=25-2\cdot6.25 \\ A=25-12.5 \\ A=12.5\text{ in}^2 \end{gathered}`