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प्रश्न
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. |
- If the length and the breadth of the rectangular field be 2x and 2y respectively, then find the area function in terms of x.
- Find the critical point of the function.
- 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.
उत्तर
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)`.
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