Question

Read the following passage and answer the questions given below.

In an elliptical sport field the authority wants to design a rectangular soccer field with the maximum possible area. The sport field is given by the graph of `x^2/a^2 + y^2/b^2` = 1.

1. If the length and the breadth of the rectangular field be 2x and 2y respectively, then find the area function in terms of x.
2. Find the critical point of the function.
3. Use First derivative Test to find the length 2x and width 2y of the soccer field (in terms of a and b) that maximize its area.
   OR
   Use Second Derivative Test to find the length 2x and width 2y of the soccer field (in terms of a and b) that maximize its area.

Answer 1

i.

Let (x, y) = `(x, b/a sqrt(a^2 - x^2))` be the upper right vertex of the rectangle.

The area function A = `2x xx 2 b/a sqrt(a^2 - x^2)`

= `(4b)/a x sqrt(a^2 - x^2), x ∈ (0, a)`.

ii. `(dA)/(dx) = (4b)/a [x xx (-x)/sqrt(a^2 - x^2) + sqrt(a^2 - x^2)]`

= `(4b)/a xx (a^2 - 2x^2)/sqrt(a^2 - x^2)`

= `-(4b)/a xx (2(x + a/sqrt(2)) (x - a/sqrt(2)))/sqrt(a^2 - x^2)`

`(dA)/(dx)` = 0

⇒ x = `a/sqrt(2)`

x = `a/sqrt(2)` is the critical point.

iii. For the values of x less than `a/sqrt(2)` and close to `a/sqrt(2), (dA)/(dx) > 0` and for the values of x greater than `a/sqrt(2)` and close to `a/sqrt(2), (dA)/(dx) < 0`.

Hence, by the first derivative test, there is a local maximum at the critical point x = `a/sqrt(2)`. Since there is only one critical point, therefore, the area of the soccer field is maximum at this critical point x = `a/sqrt(12)`

Thus, for maximum area of the soccer field, its length should be `asqrt(2)` and its width should be `bsqrt(2)`.

OR

A = `2x xx 2 b/a sqrt(a^2 - x^2), x ∈ (0, a)`

Squaring both sides, we get

Z = A² = `(16b^2)/a^2 x^2(a^2 - x^2)`

= `(16b^2)/a^2 (x^2a^2 - 4x^4),x ∈ (0, a)`

A is maximum when Z is maximum.

`(dZ)/(dx) = (16b^2)/a^2 (2xa^2 - 4x^3)`

= `(32b^2)/a^2 x(a + sqrt(2)x)(a - sqrt(2)x)`

`(dZ)/(dx)` = 0

⇒ x = `a/sqrt(2)`

`(d^2Z)/(dx^2) = (32b^2)/a^2 (a^2 - 6x^2)`

`((d^2Z)/(dx^2))_(x = a/sqrt(2)) = (32b^2)/a^2 (a^2 - 3a^2)`

= –64b² < 0

Hence, by the second derivative test, there is a local maximum value of Z at the critical point x = `a/sqrt(2)`. Since there is only one critical point, therefore, Z is maximum at x = `a/sqrt(2)`, hence, A is maximum at x = `a/sqrt(2)`. 

Thus, for the maximum area of the soccer field, its length should be `asqrt(2)` and its width should be `bsqrt(2)`.