Asked by John
                find the dimensions of a rectangle of maximum size that will fit into a half circle of radius 30 cm. 
            
            
        Answers
                    Answered by
            Reiny
            
    make a sketch using 1/2 a circle
let the base of the rectangle be 2x and the height y
draw in the radius to one of the vertices of the rectangle, giving you a right-angled triangle so that
x^2 + y^2 = 900 (can you see why I called the base 2x ? )
area = 2xy
= 2x√(900 - x^2) = 2x (900-x^2)^1/2
d(area) = 2x(1/2)(900-x^2)^(-1/2) (-2x) + 2(900-x^2)^(1/2)
= 0 for a max of area
2x(1/2)(900-x^2)^(-1/2) (-2x) + 2(900-x^2)^(1/2)
2(900-x^2)^(-1/2) [-x^2 + 900 - x^2] = 0
900 - 2x^2 = 0
x^2 = 450
x = ±√450 = √225√2
= 15√2
then in x^2 + y^2 = 900
450 + y^2 = 900
y^2 = 450
y =15√2
so the base is 30√2 and the height is 15√2
check: 30√2 * 15√2 = 900
    
let the base of the rectangle be 2x and the height y
draw in the radius to one of the vertices of the rectangle, giving you a right-angled triangle so that
x^2 + y^2 = 900 (can you see why I called the base 2x ? )
area = 2xy
= 2x√(900 - x^2) = 2x (900-x^2)^1/2
d(area) = 2x(1/2)(900-x^2)^(-1/2) (-2x) + 2(900-x^2)^(1/2)
= 0 for a max of area
2x(1/2)(900-x^2)^(-1/2) (-2x) + 2(900-x^2)^(1/2)
2(900-x^2)^(-1/2) [-x^2 + 900 - x^2] = 0
900 - 2x^2 = 0
x^2 = 450
x = ±√450 = √225√2
= 15√2
then in x^2 + y^2 = 900
450 + y^2 = 900
y^2 = 450
y =15√2
so the base is 30√2 and the height is 15√2
check: 30√2 * 15√2 = 900
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