A grain silo consists of a cylindrical concrete tower surmounted by a metal hemispherical dome. The metal in the dome costs 1.8 times as much as the concrete (per unit of surface area). If the volume of the silo is 700 m^3, what are the dimensions of the silo (radius and height of the cylindrical tower) that minimize the cost of the materials? Assume the silo has no floor and no flat ceiling under the dome. What is the function of the cost of the silo, C, in terms of the radius, r?

C=??

The interval of interest of the objective function is: ??

(Type in interval notation)^

The radius of the cylindrical base (and of the hemispherical dome) is: ??

The height of the cylindrical base is: ??

(Round to the nearest tenth as needed for the last 3 questions)

2x3 – 6x2 – 14x + 24 square centimeters

if i'm wrong i'm sorry : )

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)

30.5*10=350$

you have to multiply the number without the decimal point then add it later but if its tenn u must move the place value of the decimal point 1 to the right.