Respuesta :
Answer:
(a) The depth that will provide the maximum cross-sectional area is 7 Inches.
(b) (b) The depths between 3 inch and 11 inch will allow at least 66 square inches of water to flow.
Step-By-Step Explanation:
Let the height of the gutter =x
Let the length of the base of the gutter = y
This may illustrate an end view of the gutter: x|_y_|x
The Aluminum Sheet =28 Inches Wide
Therefore: y+2x=28
y=28-2x
(a)Cross Sectional Area of the Gutter= Length X breadth
[TeX]Area=x(28-2x)=-2x^{2}+28x[/TeX]
The Maximum of the area which is a parabola occurs when [TeX]x = -\frac{b}{2a} = -\frac{28}{2(-2)}=7[/TeX]
A gutter depth of 7 inch will provide maximum cross-sectional area and hence allow the most water to flow.
(b)If the Cross-Sectional Area = 66 square inches
[TeX]Area=-2x^{2}+28x=66 [/TeX]
[TeX]-2x^{2}+28x-66=0[/TeX]
[TeX]-2x^{2}+22x+6x-66=0[/TeX]
[TeX]-2x(x-11)+6(x-11)=0[/TeX]
(-2x+6)(x-11)=0
-2x+6=0 OR x-11=0
x=3 or x=11
The depths between 3 inch and 11 inch will allow at least 66 square inches of water to flow.
Answer:
A) Gutter is 7 inches by 14 inches and depth that will provide max area is 7 inches.
B) Between 3 inches and 11 inches
Step-by-step explanation:
First of all let's define the sides of the gutter.
Let the height of the gutter be x
Also, let the base of the gutter be y
Thus, the total length of the 3 sides will be; x + x + y = 2x + y
Since the total perimeter is 28inches. Thus, 2x + y = 28
And, y = 28 - 2x
Cross sectional area; A = xy
Thus, cross sectional area; A = x(28 - 2x) = 28x - 2x²
So, f(x) = A = -2x² + 28x
Now, x coordinate of vertex is gotten from; -b/2a
Thus, x = - (28/(-2•2)) = 7
Thus,y coordinate at this point is;
y = 28 - 2(7) = 14
Thus maximum area will occur at x = 7 and y = 14.
So maximum area = xy = 14 x 7 = 98 in²
B) We have that;
A = -2x² + 28x
Thus, if we want at least A = 66
Which means
-2x² + 28x = 66
Thus,
-2x² + 28x - 66 = 0
Solving for the roots, we get;
x = 3 inches or 11 inches
