Question

Gloria would like to construct a box with volume of exactly 55ft3 using only metal and wood. The metal costs $7/ft2 and the wood costs $5/ft2. If the wood is to go on the sides, the metal is to go on the top and bottom, and if the length of the base is to be 3 times the width of the base, find the dimensions of the box that will minimize the cost of construction. Round your answer to the nearest two decimal places.

Answer 1

Answer:

The dimensions of the box are:

W=2.97 ft

L = 8.91ft

H=2.08ft

Step-by-step explanation:

Volume of the box= 55ft³

LWH=55

Length of the base, L = 3W

3W*WH=55

3W²H=55

$H= \frac{55}{3W^2}$

Metal costs =$7/ft²

Wood costs =$5/ft².

Area of the Sides= 2(LH+WH)

=$\frac{2*3W*55}{3W^2}+\frac{2*W*55}{3W^2}\\=\frac{110}{W}+\frac{110}{3W}$

Cost of the sides therefore is:

$5(\frac{110}{W}+\frac{110}{3W})\\=\frac{550}{W}+\frac{550}{3W}$

Area of the top and bottom=2LW=2*3W*W=6W²

Cost=7*6W²=42W²

Total Cost, C(W) of the box:

$=\frac{550}{W}+\frac{550}{3W}+42W^2\\=\frac{1650+550+126W^3}{3W}\\=\frac{2200+126W^3}{3W}$

To find the minimum cost, we set the derivative of C(W) to be equal to zero.

$C^{1}W=\frac{-2200+84W^3}{3W^2}$

$\frac{-2200+84W^3}{3W^2}=0$

84W³=2200

W³=26.19

W=2.97 ft

L = 3W=3*2.97=8.91ft

$H= \frac{55}{3X2.97^2}=2.08ft$

The dimensions of the box are:

W=2.97 ft

L = 8.91ft

H=2.08ft