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.



CBSE Class 12 Sample Paper for 2023 Boards
CBSE Class 12 Sample Paper for 2023 Boards
Last updated at Dec. 13, 2024 by Teachoo
Question 37 (iii) (Choice 2) 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.Finding (𝒅^𝟐 𝒁)/(𝒅𝒙^𝟐 ) 𝑑𝑍/𝑑𝑥=(32𝑏^2)/𝑎^2 × (𝑎^2 𝑥−2𝑥^3) Differentiating w.r.t 𝑥 (𝑑^2 𝑍)/(𝑑𝑥^2 )=(32𝑏^2)/𝑎^2 × (𝑎^2−2 × 3𝑥^2) (𝒅^𝟐 𝒁)/(𝒅𝒙^𝟐 )=(𝟑𝟐𝒃^𝟐)/𝒂^𝟐 × (𝒂^𝟐−𝟔𝒙^𝟐) Putting x = 𝒂/√𝟐 (𝑑^2 𝑍)/(𝑑𝑥^2 )=(32𝑏^2)/𝑎^2 × (𝑎^2−6(𝑎/√2)^2 ) (𝑑^2 𝑍)/(𝑑𝑥^2 )=(32𝑏^2)/𝑎^2 × (𝑎^2−6 ×𝑎^2/2) (𝑑^2 𝑍)/(𝑑𝑥^2 )=(32𝑏^2)/𝑎^2 × (𝑎^2−3𝑎^2 ) (𝑑^2 𝑍)/(𝑑𝑥^2 )=(32𝑏^2)/𝑎^2 × −2𝑎^2 < 0 Since 𝐙^′′ < 0 for x = 𝒂/√𝟐 ∴ Z is maximum when x = 𝒂/√𝟐 Thus, A is maximum at x = 𝒂/√𝟐 Finding length 2x and 2y Length = 2x = 2 × 𝑎/√2 = √𝟐a Breadth = 2y = 2 × (" " 𝒃)/𝒂 √((𝒂^𝟐 − 𝒙^𝟐 ) ) = 2b/𝑎 ×√(𝑎^2−(𝒂/√𝟐)^2 ) = 2b/𝑎 ×√(𝑎^2−𝑎^2/2) = 2b/𝑎 ×√(𝑎^2/2) = 2b/𝑎 ×𝒂/√𝟐 = 2b/√𝟐 = √𝟐 𝒃