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APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF DERIVATIVES4 Marks
Question
Kyra has a rectangular painting canvas having a total area of $24\ ft^2$ which includes a border of $0.5$ ft. on the left right and a border of $0.75$ ft. on the bottom, top inside it.

Based on the above information, answer the following questions.
  1. If Kyra wants to paint in the maximum area, then she needs to maximize.
  1. Area of outer rectangle.
  2. Area of inner rectangle.
  3. Area of top border.
  4. None of these.
  1. If x is the length of the outer rectangle, then area of inner rectangle in terms of x is.
  1. $(\text{x}+3)\Big(\frac{24}{\text{x}}-2\Big)$
  2. $(\text{x}-1)\Big(\frac{24}{\text{x}}+1.5\Big)$
  3. $(\text{x}-1)\Big(\frac{24}{\text{x}}-1.5\Big)$
  4. $(\text{x}-1)\Big(\frac{24}{\text{x}}\Big)$
  1. Find the range of x.
  1. $(1, \infty)$
  2. $(1, 16)$
  3. $(-\infty, 16)$
  4. $(-1, 16)$
  1. If area of inner rectangle is maximum, then x is equal to.
  1. 2 ft.
  2. 3 ft.
  3. 4 ft.
  4. 5 ft.
  1. If area of inner rectangle is maximum, then length and breadth of this rectangle are respectively.
  1. 3 ft, 4.5 ft.
  2. 4.5 ft, 5 ft.
  3. 1 ft, 2 ft.
  4. 2 ft, 4 ft.
✓

Answer

  1. (b) Area of inner rectangle
Solution:
In order to paint in the maximum area, Kyra needs to maximize the area of inner rectangle.
  1. (c) $(\text{x}-1)\Big(\frac{24}{\text{x}}-1.5\Big)$
Solution:
Let x be the length and y be the breadth of outer rectangle.
$\therefore$ Length of inner rectangle = $x - 1$
and breadth of inner rectangle = y - 1.5
$\therefore$ A(x) = (x - 1)(y - 1.5) [$\therefore$ xy = 24 (given)
$=\big(\text{x}-1\big)\bigg(\frac{24}{\text{x}}-1.5\bigg)$
  1. (b) $(1, 16)$
Solution:
Dimensions of rectangle (outer/inner) should be positive.
$\therefore\text{x}-1>0\text{ and }\frac{24}{\text{x}}-1.5>0$
$\therefore\text{x}>1\text{ and }{\text{x}}<16$
$\therefore$ Range of x is (1, 16)
  1. (c) 4 ft
Solution:
We have $\text{A}'\text{(x)}=(\text{x}-1)\bigg(\frac{24}{\text{x}}-1.5\bigg)$
$\Rightarrow\text{A}'(\text{x)}=(\text{x}-1)\Big(\frac{-24}{\text{x}^2}\Big)+\bigg(\frac{24}{\text{x}}-1.5\bigg)$
$=\frac{24}{\text{x}^2}-1.5\ \text{and}\ \text{A}''(\text{x})=\frac{-48}{\text{x}^3}$
For A(x) to be maximum or minimum, A'(x) = 0
$\Rightarrow-1.5+\frac{24}{\text{x}^2}=0$
$\Rightarrow\text{x}^2=16$
$\Rightarrow\text{x}=\pm4$
$\therefore\text{x}=4$ [Since, length can't be negative]
Also, $\text{A}''(4)=\frac{-48}{4^3}<0$
Thus, at x = 4, area is maximum.
  1. (a) 3 ft, 4.5 ft.
Solution:
If area of inner rectangle is maximum, then Length of inner rectangle = x - 1 = 4 - 1 = 3 ft.
And breadth of inner rectangle $=\text{y}-1.5=\frac{24}{\text{x}}-1.5$
$=\frac{24}{4}-1.5=6-1.5=4.5\ \text{ft}$

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