I have been trying this question for ages.
1. An yacht club is designed its new flag. the flag consists of a red triangle on a yellow rectangular background. In the yellow rectangle ABCD, AB mesures 8 units and AD mesures 6 units. E and F lie on BC and CD, x units from B and C as shown in the diagram.
a) Show that the area, H square units, of the red triangle AEF is
given by H(x) = 24-4x+1/2
x2 (x squared)
and
b) Hence find the greatest and least possible values of the area of triangle AEF
Its in one of your h/w excersise DIFFERENTIATION HOMEWOEK TM
Please help me
Calculate the area of the rectangle and then subtract the area of the three triangles formed between rectangle and flag ie triangles ABE, FCE and ADF. Area of a triangle is 1/2 base x altitude
Area of rectangle = 8 * 6 units squared = 48 units squared
Area of triangle ABE = 4X units squared (Base AB = 8 units)
Area of triangle FCE = X/2(6-X) units squared = (3X - ½X^2) units squared
Area of triangle ADF = 3(8-X) units squared = (24 – 3X) units squared
Total area of three triangles = 24 – 3X + 3X - ½X^2 + 4X
= 24 + 4X - ½X^2
Area of flag = Area of rectangle – area of triangles
= 48 – (24 + 4X - ½X^2)
H = 24 – 4X + ½X^2 units squared
Differentiating gives dH/dX = -4 + X
therefore have a turning point which is a minimum at X = 4
Substitute X = 4 in equation for H to give minimum area of 16
X can only have values between 0 and 6, and plot of function H will show maximum area of triangle of 24 square units occurs when X = 0.
Hope this helps
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