AI HL · 12.9
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Optimisation in context
12.9.13
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An open box with a square base of side x cm and height h cm is made from 432 cm² of card. Change x. The card is fixed, so h changes too. Find the x that gives the greatest volume.
V = 108x − 0.25x³, so dV/dx = 108 − 0.75x² = 0, which gives x = 12
The box is drawn in proportion, then enlarged or reduced to fit the picture. Compare the shapes, not the sizes.
Change x to see the box and its volume.
The concentration of a drug in the blood, \(C\) mg per litre, t hours after it is taken is modelled by \(C=8te^{-0.5t}\), for \(0\le t\le 10\). Find the time when the concentration is greatest, and the greatest concentration.
Have a go first. Then open the working one line at a time.
Graph \(y=8xe^{-0.5x}\) for \(0\le x\le 10\)
From the calculator: use x for t. The graph rises, peaks, then falls. Sketch it.
From the calculator: maximum at (2, 5.89)
Use the calculator's maximum tool.
The concentration is greatest after 2 hours, at 5.89 mg per litre
Say what the numbers mean, with units.
An open box (no lid) has a square base of side x cm and height h cm. It is made from 432 cm² of card. Find the greatest possible volume of the box.
Have a go first. Then open the working one line at a time.
\(V=x^2h\) and \(S=x^2+4xh\)
The base is one square. There are four sides and no top.
\(x^2+4xh=432\), so \(h=\frac{432-x^2}{4x}\)
The area is fixed at 432. Make h the subject.
\(V=x^2\cdot\frac{432-x^2}{4x}=108x-\frac{x^3}{4}\)
Now V depends on x only.
Graph \(y=108x-\frac{x^3}{4}\), x > 0. From the calculator: maximum at (12, 864)
Only positive x makes sense for a length.
The greatest volume is 864 cm³, when x = 12 and h = 6
Check h: \(\frac{432-144}{48}=6\).
A closed cylindrical can has radius r cm and height h cm. Its volume is 400 cm³. Find the smallest possible surface area of the can.
Have a go first. Then open the working one line at a time.
\(V=\pi r^2h=400\) and \(S=2\pi r^2+2\pi rh\)
Two ends and a curved side.
\(h=\frac{400}{\pi r^2}\)
Make h the subject of the volume equation.
\(S=2\pi r^2+2\pi r\cdot\frac{400}{\pi r^2}=2\pi r^2+\frac{800}{r}\)
The \(\pi\) cancels in the second term.
Graph \(y=2\pi x^2+\frac{800}{x}\), x > 0. From the calculator: minimum at (3.99, 301)
Use the minimum tool this time.
The smallest surface area is 301 cm², when r = 3.99 cm and h = 7.99 cm
A smaller area means less metal.
A rectangular sheet of card measures 40 cm by 25 cm. A square of side x cm is cut from each corner and the sides are folded up to make an open box. Find the greatest possible volume of the box.
Have a go first. Then open the working one line at a time.
Length \(40-2x\), width \(25-2x\), height x
Two squares come off each side, so each side loses 2x.
\(V=x(40-2x)(25-2x)\), for \(0<x<12.5\)
The width \(25-2x\) must stay positive, so x is less than 12.5.
Graph \(y=x(40-2x)(25-2x)\) for 0 < x < 12.5. From the calculator: maximum at (5, 2250)
Use the maximum tool on the part that makes sense.
The greatest volume is 2250 cm³, when x = 5
Check: the box is 30 cm by 15 cm by 5 cm, and \(30\times15\times5=2250\).
A running track has two straight sides of length L metres. At each end there is a semicircle of diameter w metres. The perimeter of the track is 400 m. Find the greatest possible area of the rectangle between the two semicircles.
Have a go first. Then open the working one line at a time.
\(2L+\pi w=400\)
Two straight sides, and two semicircles that together make a circle of perimeter \(\pi w\).
\(L=\frac{400-\pi w}{2}\)
Make L the subject.
\(A=Lw=\frac{w(400-\pi w)}{2}\), for \(0<w<\frac{400}{\pi}\)
L must be positive, which limits w.
Graph \(y=\frac{x(400-\pi x)}{2}\). From the calculator: maximum at (63.7, 6370)
Use x for w on the calculator.
The greatest area is 6370 m², when w = 63.7 m and L = 100 m
Check L: \(\frac{400-\pi(63.66)}{2}=100\).
A theatre sells 150 tickets for a show at €20 each. For every €1 increase in the price, 4 fewer tickets are sold. Each ticket sold costs the theatre €8. The price must be a whole number of euros. Find the ticket price that gives the greatest profit, and the profit.
Have a go first. Then open the working one line at a time.
Let x be the number of €1 increases. Price \(=20+x\). Tickets sold \(=150-4x\)
Both the price and the sales depend on x.
\(P=(20+x-8)(150-4x)=(12+x)(150-4x)\), for \(0\le x\le 37.5\)
Profit per ticket times the number of tickets.
Graph \(y=(12+x)(150-4x)\). From the calculator: maximum at (12.75, 2450.25)
This is the best if x could be any number.
x = 12: \(P=24\times102=2448\). x = 13: \(P=25\times98=2450\)
x must be a whole number. Compare the whole numbers either side of 12.75.
Charge €33. The profit is €2450
x = 13 gives the larger profit, and the price is \(20+13=33\).