a piece of rectangular sheet metal is 20 in wide. It's to be made into rain gutter by turning the edges to form parallel sides. Let x represent the length of each of parallel sides.
a. give restrictions on x.
b. determine a function of A that gives the area of a cross section of the gutter.
c. for what values of x will A be a max (and thus maximize the amount of water that gutter will hold?
d. what is max area?
e. for what values of x will the area of a cross section be less than 40 square in?
a. To determine the restrictions on x, we need to consider the width of the sheet metal (20 inches) and the fact that turning the edges forms parallel sides.
Considering the rain gutter's cross-section, we can see that the width of the sheet metal will be divided into three segments: x inches, 20 inches, and x inches.
To form parallel sides, the sum of the two smaller segments (x inches each) must be less than or equal to the length of the longer segment (20 inches). Mathematically, this can be expressed as:
2x ≤ 20
Simplifying the inequality gives:
x ≤ 10
Therefore, the restrictions on x are x ≤ 10.
b. To determine the function of A that gives the cross-sectional area of the gutter, we need to consider the dimensions of the cross-section.
The length of the cross-section will be the sum of the three segments, which are x, 20, and x. Therefore, the length is given by:
L = 2x + 20
The height of the cross-section is the width of the sheet metal, which is 20 inches.
The area of a rectangle is given by the formula A = length × width. Substituting the values we have:
A = (2x + 20) × 20
Simplifying, the function of A is:
A = 40x + 400
c. To find the values of x that maximize the area, we can use calculus. We need to find the critical points by taking the derivative of the area function and setting it equal to zero:
dA/dx = 40
Setting dA/dx equal to zero, we have:
40 = 0
Since this equation has no solution, there are no critical points. However, we still need to check the endpoints of the interval (x ≤ 10).
When x = 10, the area function becomes:
A = 40(10) + 400 = 800 square inches
Therefore, the maximum value of A is 800 square inches.
d. The maximum area is 800 square inches.
e. To find the values of x for which the area of the cross-section is less than 40 square inches, we can set the area function less than 40 and solve for x:
40x + 400 < 40
Subtracting 400 from both sides gives:
40x < -360
Dividing both sides by 40, we get:
x < -9
Since negative lengths don't make sense in this context, there are no values of x for which the area of the cross-section is less than 40 square inches.