# A surfboard is in the shape of a rectangle and semicircle.The perimeter is 4m.Find the maximum area of the surfboard.

beckden | High School Teacher | (Level 1) Educator

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x = length of the rectangle.

y = width of rectangle

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The area of the rectangle is x*y, the semicircle has a radius of y/2 so its area is 1/2pi(y/2)^2 so the total area is

`A=xy+1/8piy^2`

The perimeter is `2x+y+pi(y/2) = 2x+y+pi/2y` .   We are told the perimeter = 4m

So 2x+y+pi/2y = 4

Solving for x we get x = 1/2(4 - (1+pi/2)y)=2 - y/2 - pi/4 y

So substituting into the equation for area

` A=(2 - y/2 - pi/4y)y+pi/2(y/2)^2 = 2y - y^2/2 -pi/4y^2 + pi/8y^2 = 2y - y^2/2 + piy^2/8`

Take the derivative with respect to y to find the extreema.

`(dA)/(dy)=2 - y - ypi/4`

`(dA)/(dy) = 0` when `2 - y - piy/4 = 0`

`(1/4pi+1)y = 2`

`y = 8/(4+pi)`

`x = 2 - 2(8/(4+pi))-2pi(8/(4+pi))=4/(4+pi)`

So the answer is length `= 4/(4+pi)` , width = `8/(4+pi)` Because obviously x=0 or y=0 do not give the maximum.  We also see (d^2A)/(dy^2)=-1-pi/4 which is negative and therefore the answer is a maximum.