A swimming pool is to be constructed with reinforced...

A swimming pool is to be constructed with reinforced concrete. The pool will have a square base with length and width $x$ and height $h$. The cost of the concrete, including labour and materials, forming the base is $p$ per square metre. The concrete for the side walls costs $q$ per square metre. The budget for the construction cost of the pool is $M$. (a) Setting up a constrained-optimization problem to maximize the volume of the pool. (b) Find the Lagrangian function. (c) Find the effect on the optimized volume with respect to a change in the cost of the base material $p$.

A swimming pool is to be constructed with reinforced concrete. The pool will have a square base with length and width $x$ and height $h$. The cost of the concrete, including labour and materials, forming the base is $p$ per square metre. The concrete for the side walls costs $q$ per square metre. The budget for the construction cost of the pool is $M$.
(a) Setting up a constrained-optimization problem to maximize the volume of the pool.
(b) Find the Lagrangian function.
(c) Find the effect on the optimized volume with respect to a change in the cost of the base material $p$.

Key Concepts

Constrained Optimization

This concept involves maximizing or minimizing a function (the objective function) while satisfying one or more constraints. In the context of the given problem, the objective is to maximize the volume of the pool, and the constraint comes from the budget limit on the cost of the materials used for construction. This requires formulating a mathematical problem where the volume is maximized subject to the total cost (a function of the dimensions and material costs) equaling the available budget.

Lagrange Multiplier Method

The Lagrange multiplier method is a strategy used to find the local maxima and minima of a function subject to equality constraints. It works by incorporating the constraints into the objective function using additional variables called Lagrange multipliers. In this problem, the Lagrangian function combines the volume function with the cost constraint, allowing for simultaneous optimization of the dimensions (x and h) and calculation of the multiplier, which also offers insights into the sensitivity of the optimal volume to changes in the cost parameters.

Sensitivity Analysis

Sensitivity analysis evaluates how the optimal value of the objective function changes in response to variations in the parameters of the problem. Within the Lagrange multiplier framework, the multiplier associated with a constraint often represents the rate at which the optimal value of the objective function will change as the constraint is slightly relaxed or tightened. In the pool problem, this analysis investigates the effect on the optimized volume when there is a change in the cost of the base material (parameter p), demonstrating the practical use of the envelope theorem in constrained optimization.

Transcript

00:01 All right, hello.

00:01 In this question we are trying to construct a rectangular tank with a square base which is xx, x times x, and it's going to be placed in the ground and catch water.

00:12 It has a height of y and we are given that the cost function is equal to 5 times x squared plus 4xy plus 10xy and we know that the volume of our vessel needs to be 20 cubic meters.

00:25 So to start i'm going to just expand this cost function since it looks like there's some like terms in there and it'll just make it a little easier to work with.

00:34 And so simplifying that down we get 5x squared plus 30xy and now we need our constraint function and our constraint function is going to be v here.

00:44 So we know that v is equal to x squared times y, the area of the base, times the height.

00:50 We want that to equal 20.

00:52 And so solving this for y since it's a little more simple will take will be 20 over x squared.

00:58 So we know that x can't be zero for this to work and so we want to just then plug that into our cost function.

01:07 So we have 5x squared plus 30x times 20 over x squared and simplifying that down we get our cost as a function of x is going to be 5x squared plus 600 over x.

01:24 And now we want to minimize this cost function.

01:26 So to do that we're going to take the first derivative which is going to be 10x minus 600 over x squared.

01:34 Noting that x cannot equal zero we're going to set this equal to zero and solve.

01:40 So we'll have 10x cubed equals 600.

01:46 X cubed then equals 60 and x will equal the cubed root of 60 which i'll write as 60 to the one third.

01:57 Just to quickly check that that is indeed a minimum i'm going to take the second derivative here and perform the second derivative test.

02:04 So we'll get 10 minus 1200 over x cubed.

02:08 This will actually be a plus there.

02:11 And so if i plug in 60 to the one third i'm going to have a positive value.

02:19 It's going to be greater than zero because there's no negatives in there...

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