Respuesta :
2x + y = 12
y = 12 - 2x
A = x * y
A = x * (12 - 2x)
A = 12x - 2x^2
Derive
dA/dx = 12 - 4x
Set dA/dx = 0
0 = 12 - 4x
4x = 12
x = 3
A = x * y
A = 12x - 2x^2
A = 12 * 3 - 2 * 3^2
A = 36 - 2 * 9
A = 36 - 18
A = 18
So 18 square inches is the largest cross-section
x = 3 , y = 6
Those are your dimensions .
16 = x * y
16 = 12x - 2x^2
2x^2 - 12x + 16 = 0
x^2 - 6x + 8 = 0
x^2 - 6x + 9 = 1
(x - 3)^2 = 1
x - 3 = -1 , 1
x = 2 , 4
2 inches or 4 inches will give you a cross-section with an area of 16
2 </= x </= 4 is the domain
y = 12 - 2x
A = x * y
A = x * (12 - 2x)
A = 12x - 2x^2
Derive
dA/dx = 12 - 4x
Set dA/dx = 0
0 = 12 - 4x
4x = 12
x = 3
A = x * y
A = 12x - 2x^2
A = 12 * 3 - 2 * 3^2
A = 36 - 2 * 9
A = 36 - 18
A = 18
So 18 square inches is the largest cross-section
x = 3 , y = 6
Those are your dimensions .
16 = x * y
16 = 12x - 2x^2
2x^2 - 12x + 16 = 0
x^2 - 6x + 8 = 0
x^2 - 6x + 9 = 1
(x - 3)^2 = 1
x - 3 = -1 , 1
x = 2 , 4
2 inches or 4 inches will give you a cross-section with an area of 16
2 </= x </= 4 is the domain
Part A;
The depth that will provide maximum cross sectional area is 3 inches
Part B;
The depths that allow up to 16 in.² to flow are depths of 4 inches and 2 inches
Part A;
The given parameters are;
The width of the sheets for the rain gutters = 12 inches
The angle at the edges of the formed rain gutter = 90°
Method;
Derive a function for the cross sectional area and find the values of the variables of the function at the maximum point by differentiation
Let y represent the depth of the rain gutter that provide maximum cross section, and let x represent the width of the rain gutter, we have;
x + 2·y = 12
∴ x = 12 - 2·y
The cross-sectional area, A = Depth × Width
∴ A = x × y = (12 - 2·y) × y = 12·y - 2·y²
At maximum cross sectional area, [tex]A_{max}[/tex], we have;
[tex]\dfrac{\mathrm{dA} }{\mathrm{d} y} = 0[/tex]
[tex]\dfrac{\mathrm{dA} }{\mathrm{d} y} = \dfrac{\mathrm{d(12 \cdot y - 2 \cdot y^2)} }{\mathrm{d} y} = 12 - 4 \cdot y = 0[/tex]
Therefore, at [tex]A_{max}[/tex], 12 - 4·y = 0
4·y = 12
y = 12/4 = 3
y = 3
Therefore;
The depth that will provide maximum cross sectional area, y = 3 inches
Part B
The depth that will allow 16 in.² of water to flow is given as follows;
A = x · y = 16
x = 12 - 2·y
∴ A = (12 - 2·y) × y = 12·y - 2·y² = 16
12·y - 2·y² = 16
dividing both sides by 2 gives;
(12·y - 2·y²)/2 = 16/2 = 8
6·y - y² = 8
y² - 6·y + 8 = 0
(y - 4)·(y - 2) = 0
y = 4 or y = 2
Therefore, 16 in.² of water is able to flow when the depth is either 4 inches or 2 inches
Learn more about maximum area here;
https://brainly.com/question/7172514
