Arectangle warehouse will have 5000 square feet of floor space and will be separated into two rectangular rooms by an interior wall. the cost of the exterior walls is $150 per linear foot and the cost of the interior wall is $100 per linear foot. find the dimensions that will minimize the cost of building the warehouse.

x = 59.4 ft

y = 84.18ft

Step-by-step explanation:

The cost of exterior walls is $150 per linear foot.

The cost of interior walls is $100 per linear foot.

xy = 5000

y = 5000/x

For the exterior walls, we have 2(x+y)(120)

For the interior wall, we have 100x

The cost function = C

C = 2(x+y)(120) + 100x

C= 240(x+y) + 100x

= 240x + 240y + 100x

= 340x + 240y

Recall that y = 5000/x

C = 340x + 240(5000/x)

C = 340x + 1200000/x

Differentiate C with respect to x

C'(x) = 340 - 1200000/x^2

= (340x^2 -1200000) / x^2

To minimize cost C'(x) = 0

(340x^2 -1200000) / x^2 = 0

340x^2 -1200000 = 0

340x^2 = 1200000

x^2 = 1200000/340

x = √1200000/340

x = 59.4 ft

Recall that y = 5000/x

y = 5000/59.4

y = 84.18 ft

d

step-by-step explanation:

d

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!