Let the side lengths be:

• Square side $= x$ m
• Hexagon side $= y$ m

Use the wire length condition

Total wire length is $20$ m.

• Perimeter of square $= 4x$
• Perimeter of regular hexagon $= 6y$

Hence, $4x+6y=20$ $x=\frac{20-6y}{4}=5-\frac{3y}{2}$

Write the total area

• Area of square $= x^2$
• Area of regular hexagon $= 6\times \frac{\sqrt3}{4}y^2=\frac{3\sqrt3}{2}y^2$

So total area, $A(y)=x^2+\frac{3\sqrt3}{2}y^2$

Substitute $x=5-\frac{3y}{2}$: $A(y)=\left(5-\frac{3y}{2}\right)^2+\frac{3\sqrt3}{2}y^2$

Differentiate to minimize

First expand: $\left(5-\frac{3y}{2}\right)^2=25-15y+\frac{9y^2}{4}$

Therefore, $A(y)=25-15y+\frac{9y^2}{4}+\frac{3\sqrt3}{2}y^2$

Differentiate: $A'(y)=-15+\frac{9}{2}y+3\sqrt3\,y$

For minimum, set $A'(y)=0$: $-15+\left(\frac{9}{2}+3\sqrt3\right)y=0$ $\left(\frac{9}{2}+3\sqrt3\right)y=15$ $y=\frac{15}{\frac{9}{2}+3\sqrt3}$

Multiply numerator and denominator by $2$: $y=\frac{30}{9+6\sqrt3}=\frac{10}{3+2\sqrt3}$

Check that it is a minimum

$A''(y)=\frac{9}{2}+3\sqrt3>0$

So the area is indeed minimum.

Compare with options

$y=\frac{10}{3+2\sqrt3}$ This matches Option D.

Comparison with stored correct answer

Stored correct answer: D

Our derived answer: D

So they agree.