Class CBSE Class 12 Mathematics Applications of Derivatives Q #704

Read the Passage

A technical company is designing a rectangular solar panel installation on a roof using 300 metres of boundary material. The design includes a partition running parallel to one of the sides dividing the area (roof) into two sections.

Let the length of the side perpendicular to the partition be $x$ metres and with parallel to the partition be $y$ metres.,
COMPETENCY BASED
APPLY
2 Marks 2025 AISSCE(Board Exam) SUBJECTIVE
Using first derivative test, calculate the maximum area the company can enclose with the 300 metres of boundary material, considering the parallel partition.
Prev Next

AI Tutor Explanation

Powered by Gemini

Step-by-Step Solution

1. Define the variables and constraints:<\/strong><\/p>

Let \( x \) be the length of the side perpendicular to the partition and \( y \) be the length of the side parallel to the partition.

The total boundary material is 300 meters, so \( 3x + 2y = 300 \).<\/p>

2. Express \( y \) in terms of \( x \):<\/strong><\/p>

From the constraint, \( 2y = 300 - 3x \), so \( y = 150 - \frac{3}{2}x \).<\/p>

3. Define the area function:<\/strong><\/p>

The area \( A \) of the rectangular installation is \( A = xy \).<\/p>

Substitute \( y \) in terms of \( x \): \( A(x) = x(150 - \frac{3}{2}x) = 150x - \frac{3}{2}x^2 \).<\/p>

4. Find the first derivative of \( A(x) \):<\/strong><\/p>

\( A'(x) = 150 - 3x \).<\/p>

5. Set the first derivative to zero and solve for \( x \):<\/strong><\/p>

\( 150 - 3x = 0 \), so \( 3x = 150 \), and \( x = 50 \).<\/p>

6. Verify that this is a maximum using the first derivative test:<\/strong><\/p>

For \( x < 50 \), \( A'(x) > 0 \) (e.g., \( A'(40) = 150 - 3(40) = 30 > 0 \)).<\/p>

For \( x > 50 \), \( A'(x) < 0 \) (e.g., \( A'(60) = 150 - 3(60) = -30 < 0 \)).<\/p>

Since the derivative changes from positive to negative at \( x = 50 \), this is a maximum.<\/p>

7. Find the corresponding value of \( y \):<\/strong><\/p>

\( y = 150 - \frac{3}{2}(50) = 150 - 75 = 75 \).<\/p>

8. Calculate the maximum area:<\/strong><\/p>

\( A = xy = 50 \times 75 = 3750 \) square meters.<\/p>

Correct Answer: 3750<\/strong>

AI generated content. Review strictly for academic accuracy.

Pedagogical Audit
Bloom's Analysis: This is an APPLY question because it requires students to apply the concepts of derivatives to solve an optimization problem in a real-world context. Specifically, they need to formulate an area function, find its derivative, and use the first derivative test to find the maximum area.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to apply a specific algorithm (first derivative test) to solve an optimization problem. This involves knowing the steps of differentiation, setting the derivative to zero, and interpreting the result to find the maximum area.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. The question assesses the student's ability to apply calculus concepts to a real-world optimization problem, which aligns with competency-based education principles. It goes beyond rote memorization and tests the application of knowledge.

More from this Chapter

VSA
Show that the function $f(x)=\frac{16\sin x}{4+\cos x}-x$, is strictly decreasing in $(\frac{\pi}{2},\pi)$
SUBJECTIVE
Write the equation for the total boundary material used in the boundary and parallel to the partition in terms of x and y..
SUBJECTIVE
Find the critical points of the area function. Use second derivative test to determine critical points at the maximum area. Also, find the maximum area.
LA
(a) The median of an equilateral triangle is increasing at the rate of 2√3 cm/s. Find the rate at which its side is increasing. OR (b) Sum of two numbers is 5. If the sum of the cubes of these numbers is least, then find the sum of the squares of these numbers.
MCQ_SINGLE
A spherical ball has a variable diameter \(\frac{5}{2}(3x+1).\) The rate of change of its volume w.r.t. x, when \(x=1\), is :
View All Questions