Question

If \(x\) and \(y\) are positive real numbers satisfying \(x + y = 52\), then the minimum possible value of \(91\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\) is:

• A. \(\frac{5103}{52}\)
• B. \(\frac{5041}{52}\)
• C. \(\frac{3051}{52}\)
• D. \(\frac{4105}{52}\)

Hint:

AM-GM Inequality principle: For a fixed sum \(S\), the product \(P\) is maximum when all terms are equal. Conversely, for a fixed product, the sum is minimum when terms are equal.

Answer 1

The Correct Option is A

Step 1: Simplify the Expression:
Let the expression be \(E\). \[ E = 91 \left(1 + \frac{1}{x}\right) \left(1 + \frac{1}{y}\right) \] Expanding the brackets: \[ E = 91 \left(1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{xy}\right) \] \[ E = 91 \left(1 + \frac{x+y}{xy} + \frac{1}{xy}\right) \] Step 2: Substitute Known Values:
Given \(x + y = 52\), substitute into the equation: \[ E = 91 \left(1 + \frac{52}{xy} + \frac{1}{xy}\right) \] \[ E = 91 \left(1 + \frac{53}{xy}\right) \] Step 3: Minimize the Expression:
To find the minimum value of \(E\), we need to minimize \(\frac{53}{xy}\), which means we must maximize the product \(xy\). For positive real numbers with a constant sum \(x + y = 52\), the product \(xy\) is maximized when \(x = y\). \[ x = y = \frac{52}{2} = 26 \] Max \(xy = 26 \times 26 = 676\). Step 4: Calculate the Minimum Value:
Substitute \(xy = 676\) into the expression for \(E\): \[ E_{\min} = 91 \left(1 + \frac{53}{676}\right) \] \[ E_{\min} = 91 \left(\frac{676 + 53}{676}\right) = 91 \left(\frac{729}{676}\right) \] We know that \(676 = 26^2 = (2 \times 13)^2 = 4 \times 13^2\) and \(91 = 7 \times 13\). \[ E_{\min} = \frac{7 \times 13 \times 729}{4 \times 13 \times 13} = \frac{7 \times 729}{4 \times 13} \] \[ E_{\min} = \frac{5103}{52} \]