3. (15 points) A cylindrical can is to be designed to hold 300 cm3 . You are asked to find the dimensions of the can that requires the least area of plate material. Assume that the can is completely closed. Ignore the thickness of the plate material. (a) (5 points) Solve the problem graphically in a contour plot. (b) (10 points) Compare your result with the one obtained using first order necessary conditions

a) Radius: 3.13 cm

Height: 9.77 cm

Surface: 122.82 cm2

b) Attached

c) To calculate the optimal point for our function, we use the property that an extreme point (maximum or minimum) can be calculated for the point in which the first derivative is equal to zero. This assumes both the function and its derivatives are continous in the range of analysis.

Step-by-step explanation:

The can has a volume of 300 cm3.

We have to minimize the surface of the can.

The surface (objective function, in this problem) can be expressed as:

The volume (restriction, in this problem) can be written as:

We can express one variable in funtion of the other:

Then, the function S becomes:

Then, we derive and equal to zero

In the attachment:

Black line: constraint (Volume=300)

Red lines: Contour lines for the Surface function (S=90, S=110, S=140)

Green line: Contour line for the optimum Surface (S=122)

ian gone lie this got me confused

step-by-step explanation:

d) 30 units^2

area of rectangle = l × b

we know,

length = 10 unit

breadth = 3 unit

applying the formula,

= l × b

= 10 × 3

= 30 unit^2

happy to

pls mark as brainliest