Differentiation - max min Type 2 TYS 1

The questions on this page focus on situations:

2. addressing areas including paddocks, etc

4. measuring the distance between two curves.

Creating 2 shapes.	1. A wire of length 30 cm is cut into two pieces. One piece is then bent to form an equilateral triangle with each side of length x cm. The other side is bent to form a circle. (i) Show that the sum of the areas of the two figures A cm² can be expressed as (ii) Find the length of wire that must be used to form the triangle so that the sum of the areas of the two figures is a minimum. Answer.Length is 18.69 cm.
Areas and paddocks.	2. A farmer has a straight river running through her property. She wishes to create a small field using the river as one side. If the farmer has 200 metres of fencing, what dimensions must she make the field so as to enclose the largest possible area? Answer.Dimensions are 50 m × 100 m.
	3. The diagram shows the graph of y = (x - a)². The point P (p, (p - a)²) lies on the curve (with 0 < p < a). The tangent at P is drawn to intersect with the x and y axes at points A and B respectively. (i) Show that the tangent to the curve at P has the equation y = (p - a)(2x - a - p). (ii) Find the coordinates of the point P, in terms of a, which will maximise the area of the triangle ABO. Answer.(ii) P is (a/3, 4a²/9).
	4. A printed page of a book is to have side margins of 1.5 cm, a top margin of 2 cm and a bottom margin of 3 cm. The printed area of the page is to be 280 cm² in area. Find the dimensions of the page (to 1 dp) if the area of the paper used is to be a minimum. Answer.Page size is 16.0 × 21.6 cm.
	5. A rectangular paper has dimensions 12 cm wide and 20 cm long as shown. The corner at P is then folded down and placed on point P₁. When this fold is made, a small triangle △KAP₁ is formed at the bottom left. Let KA = x cm. (i) Explain why KP₁ is (12 - x) cm long. (ii) Show that the area of △AKP₁ is given by . (iii) Hence show that when x is one-third the length of PA, the area of △AKP₁ is a maximum.
	6. ABCD is a square of side 10 cm. PQRS is another square with vertices on the four sides of ABCD. It is known that BQ = CR = DS = AP = x cm. (i) Find an expression for the area of PQRS in terms of x. (ii) find the value of x which will make the area of PQRS a minimum. Answer.For a minimum, x = 5 cm.
Geometric shapes.	7.
	8. The triangle PDB with fixed side lengths b and d is shown below. (i) Show that the area A of triangle PDB is given by (ii) Hence prove that the maximum area of the triangle occurs when .
	9. ABCD is a trapezium inscribed in a semi-circle of diameter 20 cm as shown. AB=CD and O is the centre of the semicircle. (ii) Show that the area of the trapezium ABCD is given by (iii) Hence find the length of BC such that the area of the trapezium ABCD is a maximum. Answer.For max area, BC = x = 10 cm.
	10. An isosceles triangle ABC is inscribed in a circle of radius 10 cm. OP = x and OP bisects AC such that AC is perpendicular to OP. Copy or trace the diagram. (i) Show that the area A of ∆ABC is given by . (ii) Show that (iii) Hence prove that the triangle with maximum area is equilateral.
	11. In the diagram above, ABCD is a rectangle with AB = 6 cm and BC = 4 cm. E and F lie on the lines DA and DC respectively so that E, B and F are collinear. Let AE = x cm and CF = y cm. (i) Show that triangles EAB and BCF are similar. (ii) Show that xy = 24. (iii) Show that the area T of triangle EDF is given by (iv) Find the height and base of the triangle EDF when it has minimum area. Justify your answer. Answer.For min area, height, h = 8 cm and base = 12 cm.
Distance between 2 curves.	12. A is on the curve y = (x – 2)² + 9 and B is on the curve y = x (6 – x). (i) Sketch the two curves. (ii) Show that the length of AB is D = 2x² – 10x + 13. (ii) Hence find the minimum length of AB. Answer.Min length of AB = 0.5 units.
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