A rectangular building is being designed to minimize heat...

A rectangular building is being designed to minimize heat loss. The east and west walls lose heat at a rate of 10 units/m $^{2}$ per day, the north and south walls at a rate of 8 units/m $^{2}$ per day, the floor at a rate of 1 unit/m $^{2}$ per day, and the roof at a rate of 5 units/m $^{2}$ per day. Each wall mustbe at least 30 $\mathrm{m}$ long, the height must be at least $4 \mathrm{m},$ and the volume must be exactly 4000 $\mathrm{m}^{3} .$ (a) Find and sketch the domain of the heat loss as a function of the lengths of the sides. (b) Find the dimensions that minimize heat loss. (Check both the critical points and the points on the boundary of the domain.) (c) Could you design a building with even less heat loss if the restrictions on the lengths of the walls were removed?

A rectangular building is being designed to minimize heat loss. The east and west walls lose heat at a rate of 10 units/m $^{2}$ per day, the north and south walls at a rate of 8 units/m $^{2}$ per day, the floor at a rate of 1 unit/m $^{2}$ per day, and the roof at a rate of 5 units/m $^{2}$ per day. Each wall mustbe at least 30 $\mathrm{m}$ long, the height must be at least $4 \mathrm{m},$ and
the volume must be exactly 4000 $\mathrm{m}^{3} .$
(a) Find and sketch the domain of the heat loss as a function of the lengths of the sides.
(b) Find the dimensions that minimize heat loss. (Check both the critical points and the points on the boundary of the domain.)
(c) Could you design a building with even less heat loss if the restrictions on the lengths of the walls were removed?

Key Concepts

Lagrange Multipliers

This is a method used to find the extrema of a function subject to equality constraints. By introducing auxiliary variables (Lagrange multipliers), the problem is transformed into one where the gradients of the objective function and the constraints are parallel, allowing for a systematic procedure to locate potential optimal points.

Critical Points and Boundary Analysis

Once the objective function and the constraints are defined, it is important to analyze both the interior (critical points) and the boundary of the feasible region. Critical points are found by setting derivatives to zero, while boundary analysis ensures that lower or higher values of the function do not occur at the edges of the feasible set. This dual approach guarantees that the global optimum is identified.

Mathematical Modeling

Mathematical modeling involves formulating a real-world problem, such as minimizing heat loss in a building, into mathematical expressions. This includes defining an objective function that quantifies heat loss as a function of design parameters and setting up constraints based on physical and practical requirements. Such models are essential for analyzing and solving engineering and scientific problems.

Constrained Optimization

This concept involves finding the best solution (such as a minimum or maximum) for an objective function while satisfying one or more restrictions or constraints. In real-world scenarios, this typically means optimizing performance while adhering to practical limitations on variables like dimensions, materials, or costs.

Feasible Region (Domain) Determination

In optimization problems, determining the domain or feasible region is essential, as it represents all the possible values of the decision variables that satisfy the imposed constraints. This step involves analyzing inequalities and equalities such as minimum or maximum dimensions and fixed-volume conditions to establish the limits within which the solution must lie.

Transcript

0:00 Hey, it's clarissa.

00:01 So there's three parts to this equation.

00:03 So for the first part, part a, we're going to make x equal to be the length of the north and south walls and y equal to the east and west, and z is the height.

00:15 So the height loss is given with this equation, h is equal to 6xy, plus 16 xz plus 20 y z and the bottom's 4 ,000.

00:34 So x, y, z has to be equal to 4 ,000.

00:40 And we substitute z is equal to 4 ,000 over xy to get the heat loss function, which is each of x comma y is equal to 6xy plus 80 ,000 over x plus 64 ,000 over y.

01:11 Since the height must be at least four, we have to make a boundary, which is c can be equal to four or has to be bigger.

01:23 It has to be bigger, and then 4 ,000 over xy has to be bigger or equal to four.

01:34 X y has to be smaller or equal to 1 ,000 and y has to be smaller or equal to a thousand over x and also each wall must be at least 30 meters long so x has to be bigger or equal to 30 and y has to be bigger or equal to 30 so our domain is equal to x comma y for x it's bigger or equal to 30 and and 30 to 1000 over x for y.

02:31 When we draw our illustration, it'll look something like this, this, this is 100 over 3 comma 30.

02:53 This is also 100 over 3 comma 30, and then this is 30.

03:03 Comma third.

03:04 So for part b this one's a bit longer.

03:11 So first what we're going to do is we're going to find our partial equations.

03:17 So partial derivatives for hx and hy and make them equal to zero.

03:22 So after we do that we get a critical point of since our hx is equal to 6y minus 80 ,000 over x square and each y is equal to 6x minus 64 ,000 for y square.

04:00 And then we're going to calculate the partial derivatives of each x and y and determine the critical point by making them equal to zero.

04:10 So once we do that, we get a critical point of 25 .54 comma 20 .44.

04:19 For 4.

04:23 For here we have line 1.

04:28 So for l1, y is equal to 30 and y is equal to 30 and it's a bound of 30 to 100 over 3 for x.

04:51 And we're going to do each of x comma 30...

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