In reference to Inferential Statistics, for 20 degrees of freedom, the least statistical t-value among the given below is:

Let $x$x = number of shirts and $y$y = number of pants.

Profit on each shirt = ₹40
Profit on each pant = ₹30

The constraints are:

• Total items: $x + y \leq 20$x+y≤20
• Minimum shirts: $x \geq 2$x≥2
• Pants at least 4 times shirts: $y \geq 4x$y≥4x

Goal: Maximize profit $P = 40x + 30y$P=40x+30y

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Since $y \geq 4x$y≥4x and $x + y \leq 20$x+y≤20:

$$x + 4x \leq 20$$x+4x≤20

$$5x \leq 20$$5x≤20

$$x \leq 4$$x≤4

Combined with $x \geq 2$x≥2, the possible values are $x \in \{2, 3, 4\}$x∈{2,3,4}

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For $x = 2$x=2:

Minimum pants: $y \geq 4(2) = 8$y≥4(2)=8

Maximum pants: $y \leq 20 - 2 = 18$y≤20−2=18

Best choice: $y = 18$y=18

Profit $= 40(2) + 30(18)$=40(2)+30(18)

$$= 80 + 540$$=80+540

$$= 620$$=620

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For $x = 3$x=3:

Minimum pants: $y \geq 4(3) = 12$y≥4(3)=12

Maximum pants: $y \leq 20 - 3 = 17$y≤20−3=17

Best choice: $y = 17$y=17

Profit $= 40(3) + 30(17)$=40(3)+30(17)

$$= 120 + 510$$=120+510

$$= 630$$=630

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For $x = 4$x=4:

Minimum pants: $y \geq 4(4) = 16$y≥4(4)=16

Maximum pants: $y \leq 20 - 4 = 16$y≤20−4=16

Best choice: $y = 16$y=16

Profit $= 40(4) + 30(16)$=40(4)+30(16)

$$= 160 + 480$$=160+480

$$= 640$$=640

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The maximum profit is ₹640.