construction of a rain gutter: a piece of rectangular sheet metal is 20 inches wide. it is to be made into a rain gutter by turning the edges to form parallel sides. let x represent the length of each of the parallel.

a) give the 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 maximum (and thus maximize the amount of the water that the gutter will hold)?
d) what is this maximum area?
e) for what values of x will the area of a cross section be less than 40 square inches?

Look at a cross-section of the gutter.

let the height be x
then the base width = 20-2x , assuming the gutter is open at the top

a) in 20-2x , 0 < x < 10
b) Area = x(20-2x)
c) Area = 20x - 2x^2
= -2(x^2 - 10x + 25 - 25 )
= -2( (x-5)^2 - 25)
= -2(x-5)^2 + 50
d) maximum area is 50 when x = 5

e) x(20-2x) < 40
-2x^2 + 20x < 40
x^2 - 10x + 20 > 0
consider x^2 - 10x + 20 = 0
x = (10 ± √20)/2
= 5 ± √5

so the area is less than 40 for
5-√5 < x < 5+√5

a) The restrictions on x are that it should be greater than 0 and less than half the width of the metal sheet (20 inches/2 = 10 inches).

b) The function for the area of a cross section of the gutter can be represented as A(x) = x * (20 - 2x).

c) To find the values of x that will maximize A, we can take the derivative of A(x) with respect to x and set it equal to zero:

A'(x) = 20 - 4x

Setting A'(x) = 0, we find:

20 - 4x = 0
4x = 20
x = 5

Therefore, x = 5 inches will maximize the area of the cross section.

d) To find the maximum area, we substitute x = 5 back into the function A(x):

A(5) = 5 * (20 - 2*5)
A(5) = 5 * (20 - 10)
A(5) = 5 * 10
A(5) = 50 square inches

So, the maximum area of the cross section is 50 square inches.

e) To find the values of x for which the area of the cross section is less than 40 square inches, we set A(x) < 40 and solve for x:

x * (20 - 2x) < 40

Dividing both sides by x (since x will be positive):

20 - 2x < 40/x

Multiplying both sides by x:

20x - 2x^2 < 40

Rearranging:

2x^2 - 20x + 40 > 0

Dividing by 2:

x^2 - 10x + 20 > 0

Using the quadratic formula or factoring, we find that this inequality has no real solutions. Therefore, there are no values of x for which the area of the cross section is less than 40 square inches.

a) To find the restrictions on x, we need to consider the properties of the rain gutter. Since the rectangular sheet metal is 20 inches wide, the length of each of the parallel sides, represented by x, needs to be smaller than or equal to half of the width to form a gutter. Therefore, the restriction on x is:

x ≤ 10 (inches).

b) To determine the area of a cross-section of the gutter, we need to consider the dimensions of the gutter. The width will remain constant at 20 inches, while the length of each of the parallel sides is represented by x. Therefore, the area of the cross-section, A, can be calculated by multiplying the width by the length, which is:
A = x * 20 (square inches).

c) To find the values of x that will maximize the area A, we need to find the maximum point of the quadratic function A = x * 20. Since x represents the length of each of the parallel sides, it cannot be negative. Therefore, we can exclude negative values.

d) To find the maximum area, we need to find the vertex of the quadratic function. The formula to find the x-coordinate of the vertex is given by x = -b / 2a, where a is the coefficient of the x^2 term and b is the coefficient of the x term.
In this case, since there is no x^2 term, a = 0 and b = 20. Plugging these values into the formula, we get:
x = -20 / 2(0) = -20 / 0 (undefined).

Since we cannot divide by zero, it implies that there is no maximum area.

e) To find the values of x for which the area of the cross-section is less than 40 square inches, we need to set up an inequality. Using the formula from part b, we can write:
x * 20 < 40.

To solve this inequality, we can divide both sides by 20, yielding:
x < 2.

Therefore, the values of x that will result in an area of less than 40 square inches are x < 2.

how did you figure c?

thank you! that helped a lot