Asked by fareha
Construct a window in the shape of a semi-circle over a rectangle.If the distance around the outside of the window is 12 feet.What dimensions will result in the rectangle having the largest possible area?
We need to find Amax
I know the circmfrence is 12
12=w+2L+a/2(pie)
I'm not sure about the equation above.
Thank you!
We need to find Amax
I know the circmfrence is 12
12=w+2L+a/2(pie)
I'm not sure about the equation above.
Thank you!
Answers
Answered by
bobpursley
in the equation above, I hope a=w/2
so the length around the top semicircle is PI*a=PI*w/2
12= w+2L+PI w/2
12=w(1+PI/2)+2L
area= wL+1/2 PI (w/2)^2
so solve for L in the perimeter equation, and then put that in for L in the area equation.
Take the derivative of area wrespect to w, set to zero, and solve for w.
Then go back and solve for L.
so the length around the top semicircle is PI*a=PI*w/2
12= w+2L+PI w/2
12=w(1+PI/2)+2L
area= wL+1/2 PI (w/2)^2
so solve for L in the perimeter equation, and then put that in for L in the area equation.
Take the derivative of area wrespect to w, set to zero, and solve for w.
Then go back and solve for L.
Answered by
fareha
thanks!!
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