Question
A long rectangular sheet of metal, 12 inches wide is to made into a rain gutter by turning up two sides at angles of 120° to the sheet. How many inches should be turned up to give the gutter its greatest capacity?
Answers
If the sides of the gutter have length x, then the maximum capacity happens when the cross-section has maximum area. That is a trapezoid.
The depth of the gutter is x sin60°
So, the trapezoid has area a = (12-2x + 2x cos60°)/2 * x sin60°
maximum area is where da/dx = 0
Now take it away. I get x=4
The depth of the gutter is x sin60°
So, the trapezoid has area a = (12-2x + 2x cos60°)/2 * x sin60°
maximum area is where da/dx = 0
Now take it away. I get x=4
oops
a = (12-2x + 12-2x+2x cos60°)/2 * x sin60°
a = (12-2x + 12-2x+2x cos60°)/2 * x sin60°
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Socratic › ... › Solving Optimization Problems
A long rectangular sheet of metal, 12cm wide, is to be made ...
The result is in cm not in inches but it doesn't matter.
Remark:
It is written:
y' = ( 12 − 3 x ) ( √3 / 2 ) = 0
It should be written:
A' = ( 12 − 3 x ) ( √3 / 2 ) = 0
Now multiply both sides by 2 / √3
12 - 3 x = 0
12 = 3x
12 / 3 = x
4 = x
x = 4
Second derivative test:
If A" ( x ) < 0 then function has a local maximum at x.
If A" ( x ) > 0 then function has a local minimum at x.
A" ( x ) = - 3 ∙ √3 / 2 = - 2.598 < 0
Function has a local maximum at x = 4
When you see list of results click on:
Socratic › ... › Solving Optimization Problems
A long rectangular sheet of metal, 12cm wide, is to be made ...
The result is in cm not in inches but it doesn't matter.
Remark:
It is written:
y' = ( 12 − 3 x ) ( √3 / 2 ) = 0
It should be written:
A' = ( 12 − 3 x ) ( √3 / 2 ) = 0
Now multiply both sides by 2 / √3
12 - 3 x = 0
12 = 3x
12 / 3 = x
4 = x
x = 4
Second derivative test:
If A" ( x ) < 0 then function has a local maximum at x.
If A" ( x ) > 0 then function has a local minimum at x.
A" ( x ) = - 3 ∙ √3 / 2 = - 2.598 < 0
Function has a local maximum at x = 4
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