A window frame is in the shape of a semicircle joined to a rectangle. Find the maximum area of a window using 300 cm of framework.
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if the radius is r the area function f(r,b) = (pi)r^2 + 2rb and 300 = (pi)r + 2(r+b) now replace b to make f a function of r and tell me
df/dr = 2(pi)r + 2rdb/dr + 2b now put in db/dr frm eqn 2
sorry the second eqn is a little off track
i have Area= 1/2(pi)^2 + 2rh
if the radius is r the area function f(r,b) = (pi)r^2 + 2rb
perimeter = pi(r) + 2h + 2r
yes
i tried to use http://tutorial.math.lamar.edu/Classes/CalcI/MoreOptimization.aspx but the conclusion is unclear
weve got the 2 functions havent we?
u have the two functions right?
Y DONT U RESPOND??
Let the breadth of the rectangle be x, radius of the semi-circle= x/2 [If I'm imagining it correctly] Area of semi-circle=(pi/8)x^2 Area of rectangle= x.y Total area= (pi/8)x^2 + xy And, 300=pi.x/2 + x + 2y We are supposed to maximize the area, so we'll substitute the value of y in eqn 1 and differentiate it dA/dx = (pi/4)x + xdy/dx + y Differentiating the other eqn we get, 0=pi/2 + 2dy/dx dy/dx = -pi/4 and 300 - (pi/2+1)x = 2y Therefore putting the values of dy/dx and y in eqn 3 dA/dx = (pi/4)x + x. -pi/4 + 150 - (pi/2+1)x/2 Now equate this to 0 to get the max area 150 = x.(pi/2+1)/2 thus x = 300/(pi/2+1)
sorry what does breadth mean?
Yes I have two equations but I cant seem to get the maximun area correct!
the length of the side which is attached to the semi-circle
it meant to be 6298.9cm^2
How did u try
i made an equation for the perimeter and the area Area= 1/2(pi)^2 + 2rh, perimeter = pi(r) + 2h + 2r got to the point where Radius = 300/4-(pi)
to get area i should be able to substitue the radius and h into area right?
plug in h into the first eqn
i dont know what h is
calculate it from the second eqn
wait h=150-(pi)/2r-r
yeah
now plug into the first one
I get a negative number! height cannot be negative
no just plug it in terms of r
what do you mean?
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