From a strip of tin 14 inches a trapezoidal gutter is to be made by bending up the sides at an angle of 45°. Find the width of the base for greatest carrying capacity.
If the folded up edges have length x, then the base is 14-2x
The area is thus
a = ((14-2x + √2 x)+(14-2x))/2 * x/√2
da/dx = (1-2√2)x + 7√2
da/dx=0 when x = 4+√2
So, the base is 14-(4+√2) = 10-√2
To find the width of the base for the greatest carrying capacity, we need to maximize the area of the trapezoidal gutter.
Let's start by visualizing the trapezoidal gutter. We know that the sides are bent at a 45° angle, which means that the top and bottom bases of the trapezoid are parallel.
Let the width of the top base be x inches.
Since the strip of tin is 14 inches long, we can divide it into three segments: the two slant sides and the top base.
The slant sides of the trapezoid will each have a length of x inches, as they are the sides that are bent up at a 45° angle. So, the total length of the two slant sides is 2x inches.
The top base (x inches) and the bottom base (which we need to find) will make up the remaining length of the strip, which is 14 - 2x inches.
To maximize the area, we need to find the value of x that gives us the largest possible area.
The area of a trapezoid is given by the formula:
A = (1/2) * (b1 + b2) * h
Where b1 and b2 are the lengths of the bases, and h is the height.
In our case, the height of the trapezoid is the same as the width of the strip, which is 14 inches.
Using the lengths of the bases we found earlier, the area can be expressed as:
A = (1/2) * (x + (14 - 2x)) * 14
Simplifying this expression, we have:
A = (1/2) * (x + 14 - 2x) * 14
= (1/2) * (14 - x) * 14
To find the maximum area, we can take the derivative of the area function with respect to x and set it equal to zero:
dA/dx = (1/2) * (-1) * 14
= -7
Setting the derivative equal to zero, we have:
-7 = 0
Since this equation has no solution, it means that the area function does not have a maximum or minimum value. Therefore, the carrying capacity of the trapezoidal gutter does not depend on the width of the top base.
In conclusion, the width of the base does not affect the carrying capacity of the gutter.
To find the width of the base for the greatest carrying capacity, we first need to understand the problem and break it down step by step.
1. Let's start by drawing a diagram to visualize the problem.
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In this diagram, we have a trapezoidal gutter with two sides bent up at a 45° angle. The height of the gutter is represented by 'h,' and we need to find the width of the base (let's call it 'x') for maximum carrying capacity.
2. The perimeter of the strip of tin must be equal to the sum of the lengths of all the sides of the trapezoidal gutter. Since the two sides are bent up at a 45° angle, the length of each side is the hypotenuse of the right triangle formed by 'h' and 'x.'
Each side length (l) of the gutter can be found using the Pythagorean theorem:
l = √(h^2 + x^2)
In this case, we have two sides, so the total length of the sides is:
2l = 2√(h^2 + x^2)
3. The sum of the lengths of the sides should be equal to the given length of the tin strip, which is 14 inches:
2√(h^2 + x^2) = 14
4. Now, let's isolate x to one side of the equation:
√(h^2 + x^2) = 7
5. Square both sides of the equation to eliminate the square root:
h^2 + x^2 = 49
6. Rearrange the equation in terms of 'x^2':
x^2 = 49 - h^2
7. To find the carrying capacity, we can multiply the width of the base (x) by the height (h). Let's call the carrying capacity 'C':
C = x * h
8. We want to find the value of 'x' that maximizes the carrying capacity. To do this, let's substitute the value of 'x^2' from the previous equation into the carrying capacity equation:
C = √(49 - h^2) * h
9. To maximize the carrying capacity, we find the derivative of 'C' with respect to 'h' and set it equal to zero:
dC/dh = 0
10. Differentiating the equation for 'C' with respect to 'h' and setting it equal to zero, we can find the value of 'h':
dC/dh = (-(49 - h^2)^(1/2)*h) / (49 - h^2)^(1/2) + h * (1/2)(49 - h^2)^(-1/2)(-2h) = 0
Simplifying further:
-h / (49 - h^2) + (-h^2) / (49 - h^2)^(1/2) = 0
11. Multiplying through by (49 - h^2) gives us:
-h^2 + h(49 - h^2)^(1/2) = 0
12. Factoring out an 'h' from the equation:
h(-h + (49 - h^2)^(1/2)) = 0
Since h cannot be zero (as it represents the height of the gutter), we ignore the h = 0 solution.
13. Solve for 'h':
-h + (49 - h^2)^(1/2) = 0
(49 - h^2)^(1/2) = h
Squaring both sides of the equation:
49 - h^2 = h^2
2h^2 = 49
h^2 = 49/2
h = √(49/2)
14. Now that we have the value of 'h,' we can substitute it back into the equation we derived earlier to find 'x':
x^2 = 49 - h^2
x^2 = 49 - 49/2
x^2 = 49/2
x = √(49/2)
Therefore, the width of the base for the greatest carrying capacity is approximately √(49/2) inches, and the height is approximately √(49/2) inches.