In an elliptical sport field we want to designs rectangular field with the maximum possible area.the sport field is given by the graph of (x^2/a^2)+(y^2/b^2)find the length 2xand width 2y of pitch

Question

Express in terms of a and b to maximize the area of the pitch.and express the area of the pitch as a function of x only

Answer 1

ellipse equation

x² /a² + y² / b² = 1

Area of the pitch = 2x * 2y = 4yx

y² / b² = 1 - x² /a²

y = ± b √1 - x²/ a², take only the positive value

y = b √1 - x²/ a²

Area = 4bx * √1 - x²/ a² = 4b √x ²- x⁴/ a²

dA/ dx = 4b [ 0.5 ( x ²- x⁴/ a² )^-1/2 ( 2x - 4 x³/ a² )

at max area dA/dx =0

4b [ 0.5 ( x ²- x⁴/ a² )^-1/2( 2x - 4 x³/ a² ) =0

( 2x - 4 x³/ a² ) / ( x ²- x⁴/ a² )^1/2 = 0

1-Zeros

2x - 4 x³/ a² = 0 ,

2x( 1 - 2 x²/ a² ) = 0

either x=0 neglected

or

1 - 2x²/ a² = 0

x = ± a/ √2 take only positive value for x, x = a/ √2

2- undefined

( x ²- x⁴/ a² )^1/2 = 0

x² ( 1 - x² / a² ) = 0,

x=0 neglected

or

1 - x² / a² = 0

x = a

Lets check the sign of the dA/dx around these two points:

dA/ dx = 4b [ 0.5 ( x ²- x⁴/ a² )^-1/2 ( 2x - 4 x³ / a² )

1- at

x< a/√2 for example x= a/2

dA/ dx = 4b [0.5(a ² /4- a⁴ /16 a² )^-1/2 ( 2a/2 - 4 a³/8 a² )

= 4b [ 0.5(3a²/16 )^-1/2 ( a/2 )]

= positive value

2- at x> a/√2 for example x= 0.9 a

dA/ dx = 4b [ 0.5 (0.81 a ²-0.65 a⁴/ a² )^-1/2 (1.8a -2.91 a³/ a² )

= negative value

So, the change of the sign of dA/dx from negative to positive around x= a/√2

means that we max Area at this x value.

y = b √1 - x²/ a², y = b √1 - a²/2 a², y = b /√2

the dimensions of rectangular pitch is

2x*2y= 2 a/√2* 2 b/ √2

Answer 2

So the ellipse equation should be = 1: x²/a² + y²/b² = 1. We can solve explicitly for y:

y = ±b/a√(a² - x²) we only need the + value for y.

A = (2x)(2y) = 4xy = 4xb/a√(a² - x²) We could use product rule for derivatives here, but it will be easier to put everything underneath the square root, then recognize to maximize that square root expression, it is sufficient to maximize the radicand (what is under the square root): A(x) = 16b(x² - x⁴/a²)^1/2

d/dx[x² - x⁴/a²] = 2x - 4/a²x³ = 0

-2x (2/a²x² - 1) = 0 x = 0 or x = a/√2

Solving for y: (a/√2)²/a² + y²/b² = 1.

y²/b² = 1/2

y = b/√2. Notice the symmetry of the solution, as well as the tendency for "squarish" rectangles to maximize area.