[Case Based] Use Second Derivative Test to find the length 2x & width

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.

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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

Question 37 (iii) (Choice 2) - CBSE Class 12 Sample Paper for 2023 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

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.

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