Anonymous

# What values can be taken on by the following expressions?

Suppose that numbers x and y satisfy:

x^2 <= y <= 4+ sqrt(4-x^2).

What values can the following two expressions take on?

1. 3*x+4*y;

2. y/(x+3).

Relevance
• Anonymous

If you write this as two separate inequalities

y≥x² - a simple parabola

and

y≤4+√[4-x²] - a section of a circle

and graph the two equations you can see the region of the plane which satisfies both equations. The resulting region has a parabola on the bottom stetching from (-2,4) to vertex (0,0) to (2,4) and a circular section stretching from (-2,4) to top (0,6) to (2,4).

You can then see that 3x+4y and y/(x+3) can take a wide range of values. To get the min and max plot 3x+4y=C and y/(x+3)=C for different values of C and find where it intersects the valid region. These shouldn't be too hard as its a straight line.

The minimum value of the line 3x+4y=C will occur when it is tangential to the parabolic section. The gradient of the line is -3/4. The gradient of the parabolic section is 2x. So 2x=-3/4 gives x=-3/8 and y = 9/64 giving a minimum value of 3x+4y = -9/16.

The maximum value of the line 3x+4y=C will occur when it is tangential to the circular section. The gradient of the line is -3/4. The gradient of the circular section is -x/sqrt(4-x²). So x/sqrt(4-x²)=-3/4 gives x=6/5 and y = 28/5 giving a minimum value of 3x+4y = 26.

The equation y/(x+3)=C can be rewritten as y=C(x+3) so it is another straight line. A bit harder as it has a variable gradient.

The minimum value of the equation y=C(x+3) will occur when it is tangential to the parabolic section (as the x-intercept is locked at x=-3) as this will give the steepest negative gradient.. The gradient of the line is C. The gradient of the parabolic section is 2x. So 2x=C. Plugging this back into the equations to find the point (x,y) of intersection gives: y=(C/2)^2 and y = C(C/2+3). Solving simultaneously gives: C = -12 or 0. Clearly -12 is nonsense as the line has x-intercept of x=-3 and a gradient of -12 will mean it doesnt intersection the region so the minimum value is C=0. Could have also found this without the simultaneous equation via common sense argueing given the vertex of the parabola and the x-intercept of the line. Note we don't actually need the value of the point (x,y) where this minimum occurs.

The maximum value of the equation y=C(x+3) will occur when it is tangential to the circular section (as the x-intercept is locked at x=-3) as this will give the steepest positive gradient.. The gradient of the line is C. The gradient of the parabolic section is -x/√[4-x²]. So -x/√[4-x²]=C.

This is a bit harder to just plug back in. We have to solve the simultaneous equations y=4+√[4-x²] and y = -x/√[4-x²] (x+3). A bit of work will give you x = 4/25 (-3-2√).

Plugging this back into C gives the messy expression: C = (6+√)/√[253-48√21] ≈ 4.23303.

If you graph these four straight lines: 3x+4y = C and y=C(x+3) for the values of C above you will see they are tangential.

-9/16 ≤ 3x+4y ≤ 26

0 ≤ y/(x+3) ≤ (6+√)/√[253-48√21] ≈ 4.23303.

• First Grade Rocks desires to coach that the expression is clever, then showing its value may well be achieved. in spite of everything, a cost of infinity works in a manner resembling what he has. utilising the certainty that for constructive values (a + b)^.5 < a^.5 + b^.5, leads iteratively to the nested roots being bounded by way of each and each term is the two^ok th root. The threes impression each and each term on the comparable value. the effect is that it is bounded, is increasing, so converges to a pair finite decrease, it is then got here across to be 4. I opt to multiply all of it the way by way of 3, so each root is of the comparable form. yet another Ramanujan gem!

• Anonymous

x^2 <= y <= 4+ sqrt(4-x^2) ; -2<=x<=2

1.

4<=y<=4+sqrt(4)

4<=y<=6

-6<=3x<=6

16<=4y<=24

-6+16<=3x+4y<=6+24

10<=3x+4y<=30

2.

1<=x+3<=5

From 1: 4<=y<=6

(4/5)<=y/(x+3)<=(6/1)

(4/5)<=y/(x+3)<=6