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Ncert Solutions Maths class 12th Maths Class 12th Application of Derivatives

89. Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum

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Vishal Baghel | Contributor-Level 10

4 months ago

Let, x and y be the two positive number

Then, x + y = 16 y = 16 - x

Let p be the sun of the cubes then

p = x³ + y³ = x³ + (16 -x)³ = x³ + (16)³-x³- 48x (16 -x)

p = 16³ + 48x²- 76 8x

So, $\frac{d p}{d x} = 96 x - 768 .$

$\Rightarrow \frac{d^{2} p}{d x^{2}} = 96 .$

At $\frac{d p}{d x} = 0$

96x - 768 = 0

$\Rightarrow x = \frac{768}{96} = 8 .$

∴at x = 8, $\frac{d^{2} p}{d x^{2}} = 96 \cdot > 0$

So, x = 8 is a point of local minima

So, y = 16 - 8 = 8

Hence, x = 8, y = 8

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y (x) = ∫? (2t² - 15t + 10)dt
dy/dx = 2x² - 15x + 10.
For tangent at (a, b), slope is m = dx/dy = 1 / (dy/dx) = 1 / (2a² - 15a + 10).
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2a² - 15a + 10 = -3
2a² - 15a + 13 = 0 (The provided solution has 2a²-15a+7=0, suggesting a different problem or a typo)
Following the image: 2a² - 15a + 7 = 0
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a = 1/2 Rejected as a > 1. So a = 7.
b = ∫? (2t² - 15t + 10)dt = [2t³/3 - 15t²/2 + 10t] from 0 to 7.
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The lengths of the sides of a triangle are 10 + x2, 10 + x2 and 20 – 2x2. If for x = k, the area of the triangle is maximum, then 3k2 is equal to:

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Let F₁(A,B,C) = $(A \land \sim B) \lor$ $[\sim C \land (A \lor B)] \lor \sim A a n d F_{2} (A, B) = (A \lor B) \lor (B \to \sim A)$ be two logical expressions. Then:

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By truth table

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From option let it be isosceles where AB = AC then

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= $\sqrt[]{r^{2} - h^{2} - r^{2} + 2 r h}$

$x = \sqrt[]{2 h r - h^{2}} . . . . . . . . (i)$

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Hence $Δ$ be equilateral having each side of length $\sqrt[]{3} r .$

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