Respuesta :
Answer:
Width of 16.49 ft and a Length of 20.59 ft
Step-by-step explanation:
Volume of the building = 5600 ft³
Volume of the building=lwh
Since, the building has a square square front
Width=Height
Therefore: Volume = w²l
w²l=5600
[TeX]l=\frac{5600}{w^2}[/TeX]
Total Surface Area of a Cuboid =2(lw+lh+wh)
Since we are ignoring the bottom of the building, total surface area of the building=Area of Back + Area of Glass Front + Area of Two Other Sides + Area of flat top
=w²+w²+2lw+lw
The materials costs $12 per ft² for the flat roof, $18 per ft² for the sides and the back, and $20 per ft² for the glass front.
Cost of the building=18w²+20w²+2(18)lw+12lw
C(l,w)=30w²+48lw
Recall earlier that we derived:
[TeX]l=\frac{5600}{w^2}[/TeX]
Therefore:
[TeX]C(w)=30w^{2}+48w(\frac{5600}{w^2})[/TeX]
[TeX]C(w)=30w^{2}+\frac{268800}{w}[/TeX]
[TeX]C(w)=\frac{30w^{3}+268800}{w}[/TeX]
The minimum costs for the material occurs at the point where the derivative equals zero.
[TeX]C^{'}(w)=\frac{60w^{3}-268800}{w^{2}}[/TeX]
Setting it equal to zero
[TeX]60w^{3}-268800=0[/TeX]
[TeX]60w^{3}=268800[/TeX]
[TeX]w^{3}=4480[/TeX]
w=16.49 ft
Recall:
[TeX]l=\frac{5600}{w^2}[/TeX]
[TeX]l=\frac{5600}{16.49^2}[/TeX]
l=20.59 ft
The dimensions that will minimize the cost of the room are: Width of 16.49 ft and a Length of 20.59 ft
