A square-based, box-shaped shipping crate is designed to have a volume of 16 cubic ft. The material used to make the base costs twice as much (per square foot) as the material in the sides, and the material used to make the top costs half as much (per square foot) as the material in the sides. What are the dimensions of the crate that minimize the cost of materials

Step-by-step explanation:

dimension

base=x, height =y, as they both are length so x, y >0

Volume of the crate= =

cost

base=b unit money /

side material = b/2 unit money /

Top view= b/4 unit money /

Total Cost= =

function to be minimized  =  ,

put

f(x) =

upon differentiating this equation

f'(x)=0

upon solving this we get

and further testing we get f"() >0 which proves that the point is local minimum.

Hence these are the dimensions of crate that minimizes the cost of material.

37/3

step-by-step explanation:

step 1: simplify both sides of the equation.

7x+8(x+

1

4

)=3(6x−9)−8

7x+(8)(x)+(8)(

1

4

)=(3)(6x)+(3)(−9)+−8(distribute)

7x+8x+2=18x+−27+−8

(7x+8x)+(2)=(18x)+(−27+−8)(combine like terms)

15x+2=18x+−35

15x+2=18x−35

tep 2: subtract 18x from both sides.

15x+2−18x=18x−35−18x

−3x+2=−35

step 3: subtract 2 from both sides.

−3x+2−2=−35−2

−3x=−37

step 4: divide both sides by -3.

−3x

−3

=

−37

−3

x=

37

3