Question

A rectangular building with a square front is to be constructed of materials that costs 12 dollars per ft^2 for the flat roof, 18 dollars per ft^2 for the sides and the back, and 20 dollars per ft^2 for the glass front. We will ignore the bottom of the building.

(a) If the volume of the building is 5,600 ft^3, what dimensions will minimize the cost of materials? (Round your answers to the nearest hundreth such that the dimensions increase from the smallest to the largest.)

Answer 1

Width of 16.49 ft and a Length of 20.59 ft

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