Arectangular storage container with an open top is to have a volume of 10 m3 . the length of its base is twice the width. material for the base costs $10 per square meter. material for the sides costs $6 per square meter. find the cost of the materials for the cheapest such container.

$81.77

Step-by-step explanation:

The area of the base is W * 2W = 2W^2

The volume of the container = Base area * heightwhich implies that

10 = 2W^2 * H ⇒ H = 10 / [ 2W^2 ] = 5 /W^2

Sothe total surface area is given byarea of the base + side area =

2W^2 + 2(5/W^2) [ W + 2W] =

2W^2 + (10/W^2}[ 3W] =

2^W^2 + 30/W

Sothe cost, C, to be minimized is this :

C = (base cost of materials)(base area) + (side cost of materials)(side area)

C = 5(2W^2) + 3(30/W)

C = 10W^2 + 90/W

Note: Calculustake the derivative of the costset to 0 and solve

So we have

C' = 20W - 90/W^2 = 0

20W = 90/W^2

W^3 = 90/20

W^3 = 9/2

W^3 = 4.5

W = (4.5)^(1/3) ≈ 1.651

Subbing this back into the cost function, the minimized cost is

C = 10[(4.5)^(1/3)] ^2 + 90/[4.5]^(1/3) ≈ $81.77

p ÷ r = tan(x)

step-by-step explanation:

given that there is a figure below shows a right triangle: a right triangle is shown with hypotenuse equal to q units and the length of the legs equal to p units and r units.

graph is missing so i will create the graph as per given information.

the angle between the legs having lengths p units and q units is y degrees. the angle between the legs having lengths r units and q units is x degrees.

now we need to find about what is p ÷ r equal to.

so we can use trigonometric ratio tan(x)=[altitude]/[base] to find the value of p ÷ r, because p is the altitude and r is the base.

p ÷ r=tan(x)

hence final answer is p ÷ r=tan(x).