Question

i need help with number 3&4 and the answer is in red i just need to know how to work the problem

(a) y (b) y=-x3 + 6x2-9x + 3 2. The sum of two nonnegative numbers is 36 (a) Find the two numbers if the differ ence of their square roots is to be as large as possible. 0 and 36 (b) Find the two numbers if the sum of their square roots is to be as large as possible. 18 and 18 3. An isosceles triangle has its vertex at the origin and its base parallel to the r-axis with the vertices above the axis on the curve y 27-r2. Find the largest area the triangle can have. 54 square units 4. An open-top rectangular box is con- structed from a 10 inch by 16 inch piece of cardboard by cutting squares of equal side length from the corners and folding up the sides. Find the dimensions of the box of largest volume and the maximum volume. width 6 inches, length 12 inches, height 2 inches, volume 144 in3 5. Find the following indefinite integrals

Answer 1

3)

I am assuming the base of the triangle will be the points (x, y) and (-x, y).

Therefore height of the triangle will be y, length of the base will be x+x = 2x .

Area = 1/2 * base * height = 1/2 * 2x * y = xy

Given that y = 27 - x^2 so this area equation becomes A = x(27 - x^2) = 27x - x^3

now A' = 27 - 3x^2 = 0 (to know the largest area)

so the critical points are x = 3 and x = -3

x = 3 gives local maximum as A''(3)<0 and x = -3 gives local minimum as A''(-3)>0.

now put x = 3 back into A = x(27 - x^2)

  A = 54 unit^2

4)

Let length of box =x, width of the box = y and height of the box = a .

[Note that the height of the box is also equal to length of the square cut out at the corners of the flat sheet from which it is constructed.]

So volume = xya .

Considering the length of the box, it is seen that [[16−x]] / 2 = a..........[1] and that [[10 - y]] / 2 = a..........[2]

By Equating [1] and [2]

[[16−x]] / 2 = [[10−y]] / 2

y = x−6........[3]

So volume of box V =x(x−6)a , but from .....[1] a=[16−x]2

Thus volume of box in terms of x, V=[x[x−6][16−x]] / 2,

V = [x^2−6x][8−x/2]

= [8x^2−(x^3)/2−48x+(6x^2)/2 ........[4]

Differentiating .....[4] wrtx

dV/dx= 16x−(3x^2)/2−48+6x = 0 for max/ min

= 22x−(3x^2)/2−48 = 0

= [x−12][x−8/3] = 0

x=12,x=8/3.

Here V''( x) = 22−3x,

V''(x=12) = -14<0 ,So x will maximise the volume of the box.

When x=12, a = (16−12)/2 = 2.

And (10−y)/2 = a => a=2 gives a value of y =6.

V=[2][6][12]=144