Question

Bill Ding plans to build a new hardware store. He buys a rectangular lot that is 50 ft by 200 ft, the 50-ft dimension being along the street. The store is to have 4000 ft2 of floor space. Construction costs $100 per linear foot for the part of the store along the street and only $80 per linear foot for the parts along the sides and back. To what dimensions should Bill build the store to minimize construction costs

Answer 1

Answer:

Length a = 59.62

Width b = 67.09

Cost = $10,734.4

Step-by-step explanation:

Solution:

Let a be the length and b be the width.

Area of the rectangle = a x b

a x b = 4000

Cost of the construction:

Cost = 100a + 80(a + 2b)

Cost = 100a + 80a + 160b

Cost = 180a + 160b     Equation of the Cost of the construction.

Now, we need a and b to calculate the minimum cost required to build a lot.

From the area = a x b

we have,

a x b = 4000

b = 4000/a

Putting this equation into the cost of the construction.

We get:

Cost = 180a + 160 x (4000/a)

Cost = 180a + 640000/a

Differentiating with respect to a, we get

$C^{'}$ = 180 - 640000/$a^{2}$

Putting $C^{'}$ = 0

180 - 640000/$a^{2}$ = 0

Taking LCM

180$a^{2}$ - 640000 = 0

Solving for a

180$a^{2}$ = 640000

$a^{2}$ = 640000/180

$a^{2}$ = 3555.55

Taking square root

a = 59.62

As we know,

b = 4000/a

putting the minimum value of a  = 59.62

b = 4000/59.62

b = 67.09

So, now we have both the dimensions a and b

Putting the values of a and b we will get the cost of the construction.

Cost = 180a + 160b

Cost = 180(59.62) + 160(67.09)

Cost = $10,734.4

Hence, $10,734.4 will be the minimum cost of the construction for Bill Ding.