You are asked to build an open cylindrical can (i. e. no top) that will hold 171.5 cubic inches. to do this, you will cut its bottom from a square of metal and form its curved side by bending a rectangular sheet of metal.(a) express the total amount of material required for the square and the rectangle in terms of r. (b) find the radius and height of the can that will minimize the total amount of material required

The volume of a cylinder is . From the figure, the circle inside the square makes up the base, while the lateral surface area of the cylinder is the rectangle.

In terms of r, the area of the circle from the square would be  since half the square's side is r and the area of a square is the square of is side.

The area of the rectangle is 2rh. From the equation of the cylinder's volume,

, thus the area of rectangle is 

Add both areas, take its derivative, then equate to 0 to find the value of r:

    (This is the total amount of material required for the square and the rectangle in terms of r)

, substitute this value to the equation for h in terms of r.

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3