Model and solve constrained optimisation problems

Two nonnegative side lengths of a rectangle satisfy x + y = 20. Which expression is its area objective?

Try every permitted duration and compare output with derivative sign

Which duration gives the greatest output in this model?

Eliminate the second variable

For A = xy with x + y = 20, write y = 20 − x. Then A(x) = x(20 − x), with nonnegative side lengths giving 0 ≤ x ≤ 20. Zero-area endpoints are included in this mathematical comparison

1. 1Separate objective and constraint Name the quantity to maximise or minimise. Use the given relation to remove other variables; differentiating the constraint itself does not optimise the objective
2. 2State the valid domain Use physical conditions and every additional restriction. A stationary input outside this domain cannot be selected

Given

\(\displaystyle x+y=20,\quad x=6\)

If x = 6 cm, recover the other side length in centimetres

Given

\(\displaystyle A=xy,\quad x=6,\quad y=14\)

Find the area in square centimetres at that input

Stationary is a candidate, not the conclusion

Differentiate the one-variable area. Solve A′(x) = 0, retain the input if it is permitted, then evaluate its area and both endpoint areas

1. 1Compare all admissible candidates On a closed interval, evaluate every stationary input inside the interval and both included endpoints. Compare exact objective values

Given

\(\displaystyle 20-2x=0\)

Solve the stationary equation

Now require 12 ≤ x ≤ 16. Can the previous stationary input x = 10 be chosen?

A restriction can move the optimum to an endpoint

For 12 ≤ x ≤ 16, the derivative 20 − 2x is negative throughout the interval. Compare A(12) and A(16); the greater value is attained at the included left endpoint

1. 1State the valid domain Use physical conditions and every additional restriction. A stationary input outside this domain cannot be selected
2. 2Compare all admissible candidates On a closed interval, evaluate every stationary input inside the interval and both included endpoints. Compare exact objective values

Given

\(\displaystyle A(x)=x(20-x),\quad 12\le x\le16,\quad A(12)=12(20-12)\)

Find the constrained maximum area in square centimetres

An open box gives a cubic objective

Cut squares of side x from the corners of an 18 cm square and fold the sides. The box height is x and each base side is 18 − 2x. Positive dimensions require 0 < x < 9; the limiting endpoint volumes are zero. The interior maximum occurs at x = 3.

1. 1Separate objective and constraint Name the quantity to maximise or minimise. Use the given relation to remove other variables; differentiating the constraint itself does not optimise the objective
2. 2Use the right geometric model A closed square-base box has surface area 2x² + 4xH and volume x²H. An open cut-out box instead has a different surface constraint and cubic volume

Given

\(\displaystyle V(3)=3(18-2\cdot3)^2\)

Find that box volume in cubic centimetres

Which pair distinguishes an optimiser from an optimum?

A closed box has a different constraint

For a closed square-base box with surface area 96 cm², 2x² + 4xH = 96. Eliminate H before differentiating the volume. Suppose manufacturing permits 2 ≤ x ≤ 6; the recovered height is positive throughout this interval

1. 1Separate objective and constraint Name the quantity to maximise or minimise. Use the given relation to remove other variables; differentiating the constraint itself does not optimise the objective
2. 2Use the right geometric model A closed square-base box has surface area 2x² + 4xH and volume x²H. An open cut-out box instead has a different surface constraint and cubic volume

Given

\(\displaystyle V(4)=24\cdot4-\frac12\cdot4^3\)

Find the volume at x = 4 cm

Given

\(\displaystyle 24-\frac32x^2=0,\quad x>0\)

Find the positive stationary input

Recheck an additional manufacturing bound

For a closed square-base box with surface area 96 cm², suppose the permitted base side is 5 ≤ x ≤ 6. The stationary input 4 is excluded. Volume decreases throughout this interval, so the largest allowed volume occurs at x = 5. Recover the matching height from the surface-area equation

1. 1State the valid domain Use physical conditions and every additional restriction. A stationary input outside this domain cannot be selected
2. 2Recover the original quantities The optimiser is an input, while the optimum is an output. Substitute back into the original constraint for any other requested dimensions and state their units

Given

\(\displaystyle H=\frac{96-50}{20}\)

Recover the height in centimetres at x = 5.

Given

\(\displaystyle V(x)=24x-\frac12x^3,\quad 5\le x\le6,\quad V(5)=5^2\cdot\frac{23}{10}\)

Find the constrained maximum volume in cubic centimetres

A closed square-base box has surface area 96 cm² and permitted base side 5 ≤ x ≤ 6. Which report gives its maximum volume and corresponding dimensions?

Choose the objective that matches the question

Other models minimise a sum of squares or squared distance. For a nonnegative distance D, the inputs minimising D² also minimise D. In every case, compare admissible candidate values and report the requested quantity with its own units

1. 1Recover the original quantities The optimiser is an input, while the optimum is an output. Substitute back into the original constraint for any other requested dimensions and state their units
2. 2Squared distance preserves the minimiser Distance is nonnegative and square root is increasing. Minimising squared distance gives the same input as minimising distance, while avoiding a square root during differentiation

Exact volume

\(\displaystyle V=\frac{115}{16}\text{ cm}^3\)

A different box has the exact volume shown. Report this volume to exactly two decimal places