Consider the following problem: a box with an open top is to be constructed from a square piece of cardboard, 3 feet wide, by cutting out a square from each of the four corners and bending up the sides. Find the largest volume that such a box can have. (a) Draw several diagrams to illustrate the situation, some short boxes with large bases and some tall boxes with small bases. Find the volume of each configuration. Does it appear that there is a maximum volume

) See annex

b) See annex

x  =  0,5 ft

y =  2 ft   and

V = 2 ft³

Step-by-step explanation:  See annex

c) V = y*y*x

d-1) y = 3 - 2x

d-2) V = (3-2x)* ( 3-2x)* x   ⇒ V = (3-2x)²*x    

V(x) =( 9 + 4x² - 12x )*x    ⇒   V(x) = 9x + 4x³ - 12x²

Taking derivatives

V¨(x) = 9 + 12x² - 24x

V¨(x) = 0              ⇒   12x² -24x +9 = 0     ⇒  4x² - 8x + 3 = 0

Solving for x (second degree equation)

x =[ -b ± √b²- 4ac ] / 2a

we get    x₁  =  1,5       and    x₂ =  0,5

We look at y = 3 - 2x    and see that the value x₂ is the only valid root

then

x  =  0,5 ft

y =    2 ft   and

V = 0,5*2*2

V = 2 ft³

12cm

the answer is d

37/3

step-by-step explanation:

step 1: simplify both sides of the equation.

7x+8(x+

1

4

)=3(6x−9)−8

7x+(8)(x)+(8)(

1

4

)=(3)(6x)+(3)(−9)+−8(distribute)

7x+8x+2=18x+−27+−8

(7x+8x)+(2)=(18x)+(−27+−8)(combine like terms)

15x+2=18x+−35

15x+2=18x−35

tep 2: subtract 18x from both sides.

15x+2−18x=18x−35−18x

−3x+2=−35

step 3: subtract 2 from both sides.

−3x+2−2=−35−2

−3x=−37

step 4: divide both sides by -3.

−3x

−3

=

−37

−3

x=

37

3