An agricultural field is in the form of a rectangle having length X1 meters and breadth X2 meters (X1 and X2 are variable). If X1 + X2 = 40 meters,…

Correct Option (1)

The problem defines an agricultural field as a rectangle with length $X_{1}$X1​ meters and breadth $X_{2}$X2​ meters. A condition is provided that $X_{1} + X_{2} = 40$X1​+X2​=40 meters. The objective is to determine the maximum possible area of this field, which is given by $A = X_{1} \times X_{2}$A=X1​×X2​.

To find the maximum product of two positive numbers given their constant sum, the principle of the Arithmetic Mean-Geometric Mean (AM-GM) inequality is applied. The AM-GM inequality states that for any two non-negative real numbers, the arithmetic mean is always greater than or equal to the geometric mean, with equality holding when the numbers are equal.

• Arithmetic Mean (AM) = $\frac{X_{1} + X_{2}}{2}$2X1​+X2​​
• Geometric Mean (GM) = $\sqrt{X_{1}X_{2}}$X1​X2​​
• According to AM-GM: $\frac{X_{1} + X_{2}}{2} \geq \sqrt{X_{1}X_{2}}$2X1​+X2​​≥X1​X2​​

Substituting the given sum $X_{1} + X_{2} = 40$X1​+X2​=40:

$\frac{40}{2} \geq \sqrt{X_{1}X_{2}}$240​≥X1​X2​​

$20 \geq \sqrt{X_{1}X_{2}}$20≥X1​X2​​

Squaring both sides of the inequality to isolate the area $X_{1}X_{2}$X1​X2​:

$20^2 \geq X_{1}X_{2}$202≥X1​X2​

$400 \geq X_{1}X_{2}$400≥X1​X2​

This inequality indicates that the area $X_{1}X_{2}$X1​X2​ will not exceed 400 sq. m. The maximum area is achieved when $X_{1} = X_{2}$X1​=X2​. Given $X_{1} + X_{2} = 40$X1​+X2​=40, it follows that $X_{1} = X_{2} = \frac{40}{2} = 20$X1​=X2​=240​=20 meters.

Therefore, the maximum area of the field is $20 \times 20 = 400$20×20=400 sq. m.

Incorrect Options:

Options 2 (300 sq. m), 3 (200 sq. m), and 4 (80 sq. m) are incorrect because they represent values less than the maximum possible area of 400 sq. m. The question asks for the value that the area "will not exceed," which refers to the absolute upper limit or the maximum possible area. While these values are achievable for specific dimensions of $X_{1}$X1​ and $X_{2}$X2​ that sum to 40, they do not constitute the maximum possible area.

• An area of 300 sq. m is possible, for example, if $X_{1} = 10$X1​=10 m and $X_{2} = 30$X2​=30 m (sum = 40, product = 300).
• An area of 200 sq. m is possible if $X_{1} = (20 - \sqrt{200})$X1​=(20−200​) m and $X_{2} = (20 + \sqrt{200})$X2​=(20+200​) m (sum = 40, product = 200).
• An area of 80 sq. m is possible if $X_{1} = (20 - \sqrt{320})$X1​=(20−320​) m and $X_{2} = (20 + \sqrt{320})$X2​=(20+320​) m (sum = 40, product = 80).

Since these values can be attained, they are not the ceiling that the area cannot surpass. The calculated maximum area of 400 sq. m is the correct upper bound.