Let a, b, c and d be positive real numbers such that a + b + c + d = 11. If the maximum value of \(a^5b ^3c ^2d\) is \(3750\;\beta\), then the value of \(\beta\) is
- 110
- 90
- 55
- 108
The Correct Option is B
Solution and Explanation
To maximize the given expression, we use the Arithmetic Mean (A.M.) and Geometric Mean (G.M.) inequality:
A.M. ≥ G.M.
Let us assume the terms as:
\[ \frac{a}{5}, \frac{a}{5}, \frac{a}{5}, \frac{a}{5}, \frac{a}{5}, \frac{b}{3}, \frac{b}{3}, \frac{b}{3}, \frac{c}{2}, \frac{c}{2}, d \]
Now applying A.M. ≥ G.M., we get:
\[ \frac{\frac{a}{5} + \frac{a}{5} + \frac{a}{5} + \frac{a}{5} + \frac{a}{5} + \frac{b}{3} + \frac{b}{3} + \frac{b}{3} + \frac{c}{2} + \frac{c}{2} + d}{11} \geq \sqrt[11]{a^5b^3c^2d} \]
Given \(a + b + c + d = 11\), the left-hand side simplifies to:
\[ \frac{11}{11} \geq \sqrt[11]{a^5b^3c^2d} \]
Therefore:
\[ a^5b^3c^2d \leq 5^5 \cdot 3^3 \cdot 2^2 \cdot 1 \]
Calculating the maximum value:
\[ a^5b^3c^2d \leq 5^5 \cdot 3^3 \cdot 2^2 = 337500 \]
We can express this as:
\[ 337500 = 90 \cdot 3750 \quad \text{where} \, \beta = 90 \]
Explanation
To achieve the maximum value, the numbers must be distributed in the ratios consistent with their powers in \(a^5b^3c^2d\), ensuring the product is maximized. Using A.M. ≥ G.M. is crucial in such optimization problems.
Top JEE Main Mathematics Questions
- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
- If , then the general value of is:
- The general solution of
- If the constant term in the binomial expansion of $\left(\frac{x^{\frac{3}{2}}}{2}-\frac{4}{x^l}\right)^9$ is $-84$ and the coefficient of $x^{-3 l}$ is $2^\alpha \beta$, where $\beta<0$ is an odd number, then $|\alpha l-\beta|$ is equal to_____
Top JEE Main Maxima and Minima Questions
- Let the set of all positive values of \( \lambda \), for which the point of local minimum of the function \((1 + x (\lambda^2 - x^2)) \frac{x^2 + x + 2}{x^2 + 5x + 6} < 0\) be \((\alpha, \beta)\).
Then \( \alpha^2 + \beta^2 \) is equal to ________. - If the function \(f(x) = 2x^3 - 9ax^2 + 12a^2x + 1, \, a>0\) has a local maximum at \(x = \alpha\) and a local minimum at \(x = \alpha^2\), then \(\alpha\) and \(\alpha^2\) are the roots of the equation:
- The number of critical points of the function $f(x) = (x - 2)^{2/3}(2x + 1)$ is:
- Let $f(x) = 4\cos^3 x + 3\sqrt{3} \cos^2 x - 10$. The number of points of local maxima of $f$ in interval $(0, 2\pi)$ is:
- Let \( f(x) = 3\sqrt{x - 2} + \sqrt{4 - x} \) be a real-valued function. If \( \alpha \) and \( \beta \) are respectively the minimum and the maximum values of \( f \), then \( \alpha^2 + 2\beta^2 \) is equal to:
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,