1)  If 1900 square centimeters of material is available to make a box with a square base and an open top, find the largest possible volume of the box.  2) A rancher wants to fence in an area of...

1)  If 1900 square centimeters of material is available to make a box with a square base and an open top, find the largest possible volume of the box. 

2) A rancher wants to fence in an area of 2500000 square feet in a rectangular field and then divide it in half with a fence down the middle parallel to one side. What is the shortest length of fence that the rancher can use?

3)  Find the point on the line -6x+5y-3=0 which is closest to the point (4,0).

4) A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola . What are the dimensions of such a rectangle with the greatest possible area? 

Width=

Height=

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justaguide | College Teacher | (Level 2) Distinguished Educator

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1. 1900 square centimeters of material is available to make a box with a square base and an open top, the largest possible volume of the box has to be determined.

Let the length of each side of the base be L and the height of the box be H. The area of material required to make the box in terms of L and H is L^2 + 4*L*H.

This is equal to 1900. `L^2 + 4*H*L = 1900`

=> `H = (1900 - L^2)/(4*L)`

The volume of the box is `V = L^2*H` = `L^2*(1900 - L^2)/(4*L)` = `(1900*L - L^3)/4`

To maximize V, solve `(dV)/(dL) = 0`

=> `1900/4 - 3*L^2/4 = 0`

=> `3L^2 = 1900`

=> `L^2 = 1900/3`

If L = 1900, the maximum volume of the box is `((1900/3)*(1900 - 1900/3))/(4*sqrt(1900/3)) ~~` 7969 cm^3

2. A rancher wants to fence in an area of 2500000 square feet in a rectangular field and then divide it in half with a fence down the middle parallel to one side. Let the length of the side, parallel to which the field is divided in the middle be L and the other side be H. The area of the field is 2500000, H = 2500000/L

The length of fence is given by `F = 3*L + 2*2500000/L`

To minimize the length the value of L is determined by solving `(dF)/(dL) = 0`

=> `3 - 5000000/L^2 = 0`

=> `L = sqrt(5000000/3)`

The length of the fence is `3*sqrt(5000000/3) + 5000000/sqrt(5000000/3) ~~ 7746` feet.

3. The point on the line -6x+5y-3=0 which is closest to the point (4,0) has to be determined. The slope of the line perpendicular to the line -6x+5y-3=0 is -5/6. The equation of a line with slope -5/6 and passing through (4,0) is y/(x - 4) = -5/6

The point of intersection of this line with -6x+5y-3=0 is the required point

(-5/6)*(x - 4) = (6x + 3)/5

=> (-25/6)*x + 100/6 = (6x + 3)

=> 100/6 - 3 = (25/6+6)*x

=> 82/61

y = 135/61

The coordinates of the point on the line -6x+5y-3=0 closest to (4,0) is (82/61,135/61)

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