SATHEE: Application of Derivatives 4 Question 7

Application of Derivatives 4 Question 7

####8. The maximum area (in sq. units) of a rectangle having its base on the $X$ -axis and its other two vertices on the parabola, $y = 12 - x^{2}$ such that the rectangle lies inside the parabola, is

(2019 Main, 12 Jan I)

(a) 36

(b) $20 \sqrt{2}$

(d) $18 \sqrt{3}$

Show Answer

Answer:

Correct Answer: 8. (c)

Solution:

Equation of parabola is given, $y = 12 - x^{2}$

or $x^{2} = - (y - 12)$ .

Note that vertex of parabola is $(0, 12)$ and its open downward.

Let $Q$ be one of the vertices of rectangle which lies on parabola. Then, the coordinates of $Q$ be $(a, 12 - a^{2})$

Then, area of rectangle $P Q R S$

$= 2 \times ($ Area of rectangle $P Q M O)$

[due to symmetry about $Y$ -axis]

$= 2 \times [a (12 - a^{2})] = 24 a - 2 a^{3} = Δ (let) .$

The area function $Δ_{d}$ will be maximum, when

$\begin{array}{cc} \frac{d Δ}{d a} = 0 \\ \Rightarrow 24 - 6 a^{2} = 0 \\ \Rightarrow a^{2} = 4 \Rightarrow a = 2 \\ So, maximum area of rectangle \\ PQRS = (24 \times 2) - 2 (2)^{3} \\ = 48 - 16 = 32 sq units \end{array}$

Application of Derivatives 4 Question 7

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Application of Derivatives 4 Question 7

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu