A rectangular window is surmounted by an equilateral triangle. Given that the perimeter is 16 cm. Find the width of the window so that the maximum amount of light may enter.

Hello student,
Please find the answer to your question below
Let L be the height of the rectangular part of the window and
let w be the width of the window.
The perimeter of the equilateral triangle, minus the base, is 2w and the perimeter of the rectangular part of the window, minus the top, is 2L + w. The perimeter of the window will therefore be
2L + 3w = 16
The area of the triangular part of the window will be
½bh
= ½*w*w√3/2
= w²√3/4
The area of the rectangular part will be Lw, for a total area of
A = w²√3/4 + Lw
but we can get rid of L by solving the perimeter equation for L and making the substitution:
L = 6 − 3/2w
so A = w²√3/4 + (6 − 3/2w)w
= (√3−6)/4w² + 6w
We could use calculus to maximize A, but in this case we have a quadratic equation in w so we only need to find the vertex of the parabola. If you have a quadratic of the form
ax² + bx + c, the x coordinate of the vertex will be at -b/(2a)
In this example, L will be a maximum when
w = -6 / [2*(√3−6)/4]
= 12 / (6 − √3)
or, to rationalize the denominator and simplify,
w = 3(6+√3)/11
So w=3.46m