(5) Note: This is the same bonus problem as Lesson 12. This time, you have to use the method of Lagrange Multipliers to solve it. A rectangular tank with a bottom and sides but no top is to have volume 500 cubic feet. Determine the dimensions (length, width, height) with the smallest possible surface area.

The dimensions (length, width, height) with the smallest possible surface area are

10 ft, 10 ft and 5 ft respectively.

Step-by-step explanation:

The box is an open-at-the-top box.

If the length, width and height of the box are x, z and y respectively,

The surface area of a box of dimension x, y and z open at the top is given by

S(x,y,z) = 2xy + 2yz + xz

We're to minimize this function subject to the constraint that

Volume = 500 ft³

xyz = 500

The constraint can be rewritten as

xyz - 500 = 0

Using Lagrange multiplier, we then write the equation in Lagrange form

Lagrange function = Function - λ(constraint)

where λ = Lagrange factor, which can be a function of x, y and z

L(x,y,z) = 2xy + 2yz + xz - λ(xyz - 500)

We then take the partial derivatives of the Lagrange function with respect to x, y, z and λ. Because these are turning points, each of the partial derivatives is equal to 0.

(∂L/∂x) = 2y + z - λyz = 0

λ = (2y + z)/yz = (2/z) + (1/y)

(∂L/∂y) = 2x + 2z - λxz = 0

λ = (2x + 2z)/xz = (2/z) + (2/x)

(∂L/∂z) = x + 2y - λxy = 0

λ = (x + 2y)/xy = (1/y) + (2/x)

(∂L/∂λ) = xyz - 500 = 0

We can then equate the values of λ from the first 3 partial derivatives and solve for the values of x, y and z

(2/z) + (1/y) = (2/z) + (2/x)

(1/y) = (2/x)

y = (x/2)

Also,

(2/z) + (2/x) = (1/y) + (2/x)

(2/z) = (1/y)

z = 2y = 2(x/2) = x

Hence, at the point where the box has minimal surface area,

y = (x/2)

z = x

Putting these into the constraint equation or the solution of the fourth partial derivative,

xyz - 500 = 0

(x)(x/2)(x) = 500

x³ = 1000

x = 10 ft

y = (x/2) = 5 ft

z = x = 10 ft.

Hope this Helps!!!

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

step-by-step explanation:

given   that triangle lno is congruent to triangle lnm.

we have to fill up the blanks in the given question.

according to above order of congruency   kept in mind we can say that

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

ln ≅ ln due to common in both.

hence, by sas congruency,

the side angle side rule   states that if two sides and the included angle of one triangle are congruent to another triangle, then these two triangles are congruent.

77

step-by-step explanation: