Question

Let a, b, c and d be positive real numbers such that a + b + c + d = 11 If the maximum value of a ^ 5 * b ^ 3 * c ^ 2 * d is 6250 ẞ, then the value of ẞ is

Answer 1

54

Answer 2

To find the maximum value of the expression $a^5 b^3 c^2 d$ subject to the condition $a+b+c+d=11$, where $a, b, c, d$ are positive real numbers, we can use the AM-GM inequality.

The AM-GM inequality states that for $n$ non-negative numbers $x_1, x_2, \ldots, x_n$, the arithmetic mean is greater than or equal to the geometric mean: $\frac{x_1 + x_2 + \ldots + x_n}{n} \ge \sqrt[n]{x_1 x_2 \ldots x_n}$ Equality holds when $x_1 = x_2 = \ldots = x_n$.

We want to maximize the product $P = a^5 b^3 c^2 d$. The sum of the powers in the product is $5+3+2+1=11$. This matches the sum $a+b+c+d=11$. To apply AM-GM effectively, we consider 11 terms such that their sum is $a+b+c+d=11$ and their product is proportional to $a^5 b^3 c^2 d$. We split $a$ into 5 equal parts, $b$ into 3 equal parts, $c$ into 2 equal parts, and $d$ into 1 part. Let the 11 terms be: Five terms of $\frac{a}{5}$ Three terms of $\frac{b}{3}$ Two terms of $\frac{c}{2}$ One term of $\frac{d}{1}$

The sum of these 11 terms is: $5 \left(\frac{a}{5}\right) + 3 \left(\frac{b}{3}\right) + 2 \left(\frac{c}{2}\right) + 1 \left(\frac{d}{1}\right) = a+b+c+d$ Given $a+b+c+d=11$, the sum of these 11 terms is 11.

Now, applying the AM-GM inequality to these 11 terms: $\frac{\left(\frac{a}{5}\right) + \ldots + \left(\frac{a}{5}\right) + \left(\frac{b}{3}\right) + \ldots + \left(\frac{b}{3}\right) + \left(\frac{c}{2}\right) + \left(\frac{c}{2}\right) + \left(\frac{d}{1}\right)}{11} \ge \sqrt[11]{\left(\frac{a}{5}\right)^5 \left(\frac{b}{3}\right)^3 \left(\frac{c}{2}\right)^2 \left(\frac{d}{1}\right)^1}$ Substituting the sum: $\frac{11}{11} \ge \sqrt[11]{\frac{a^5 b^3 c^2 d}{5^5 \cdot 3^3 \cdot 2^2 \cdot 1^1}}$ $1 \ge \sqrt[11]{\frac{a^5 b^3 c^2 d}{5^5 \cdot 3^3 \cdot 2^2}}$ Raising both sides to the power of 11: $1^{11} \ge \frac{a^5 b^3 c^2 d}{5^5 \cdot 3^3 \cdot 2^2}$ $1 \ge \frac{a^5 b^3 c^2 d}{3125 \cdot 27 \cdot 4}$ The maximum value of $a^5 b^3 c^2 d$ is $5^5 \cdot 3^3 \cdot 2^2$. Let's calculate this value: $5^5 = 3125$ $3^3 = 27$ $2^2 = 4$ Maximum value $= 3125 \cdot 27 \cdot 4 = 3125 \cdot 108$. $3125 \times 108 = 3125 \times (100 + 8) = 312500 + (3125 \times 8) = 312500 + 25000 = 337500$.

The maximum value of $a^5 b^3 c^2 d$ is $337500$. The problem states that the maximum value is $6250 \beta$. So, we have the equation: $6250 \beta = 337500$ Now, we solve for $\beta$: $\beta = \frac{337500}{6250}$ $\beta = \frac{33750}{625}$ To simplify the fraction, we can divide both numerator and denominator by common factors. Both are divisible by 25: $\beta = \frac{33750 \div 25}{625 \div 25} = \frac{1350}{25}$ Again, both are divisible by 25: $\beta = \frac{1350 \div 25}{25 \div 25} = \frac{54}{1}$ $\beta = 54$

The equality in AM-GM holds when all the terms are equal: $\frac{a}{5} = \frac{b}{3} = \frac{c}{2} = \frac{d}{1} = k$ (for some constant $k$) So, $a=5k, b=3k, c=2k, d=k$. Substituting these into the sum $a+b+c+d=11$: $5k+3k+2k+k = 11$ $11k = 11$ $k = 1$ Thus, the maximum value occurs when $a=5, b=3, c=2, d=1$.

The final answer is $\boxed{54}$.

Explanation of the solution:

1. Identify the expression to maximize ($P = a^5 b^3 c^2 d$) and the constraint ($a+b+c+d=11$).
2. Recognize that AM-GM inequality is suitable for maximizing a product given a constant sum.
3. To apply AM-GM, create terms whose sum is constant and whose product yields the desired expression. Since the powers are 5, 3, 2, 1, consider 5 terms of $a/5$, 3 terms of $b/3$, 2 terms of $c/2$, and 1 term of $d/1$.
4. The sum of these $5+3+2+1=11$ terms is $(a/5)\times 5 + (b/3)\times 3 + (c/2)\times 2 + (d/1)\times 1 = a+b+c+d = 11$.
5. Apply AM-GM: $\frac{\text{sum of terms}}{\text{number of terms}} \ge \sqrt[\text{number of terms}]{\text{product of terms}}$. $\frac{11}{11} \ge \sqrt[11]{\left(\frac{a}{5}\right)^5 \left(\frac{b}{3}\right)^3 \left(\frac{c}{2}\right)^2 \left(\frac{d}{1}\right)^1}$.
6. Simplify and solve for the maximum value of $a^5 b^3 c^2 d$: $1 \ge \frac{a^5 b^3 c^2 d}{5^5 \cdot 3^3 \cdot 2^2}$, so $a^5 b^3 c^2 d \le 5^5 \cdot 3^3 \cdot 2^2$.
7. Calculate the maximum value: $5^5 \cdot 3^3 \cdot 2^2 = 3125 \cdot 27 \cdot 4 = 337500$.
8. Equate this maximum value to the given form $6250 \beta$: $6250 \beta = 337500$.
9. Solve for $\beta$: $\beta = \frac{337500}{6250} = 54$.