A box with a square base and open top must have a volume of 171500 c m 3 . We wish to find the dimensions of the box that minimize the amount of material used. First, find a formula for the surface area of the box in terms of only x , the length of one side of the square base. [Hint: use the volume formula to express the height of the box in terms of x .] Simplify your formula as much as possible.

The dimensions of the box that minimize the amount of material used is length of square base x = 70 cm and height of box = 35 cm

Step-by-step explanation:

The volume of the box with square base is V = x²h where x = length of side of square base and h = height of box.

Now, the area of the of box open at the top with dimensions of length, l , width, w and height, h are

A = 2lh + lw + 2wh

Now for a box of square base, l = w = x.

So, A = 2xh + x² + 2xh

A = x² + 4xh

Since V = x²h and V = 171500 cm³,

h = V/x² = 171500/x²

substituting h into A, we have

A =   x² + 4xh

A =  x² + 4x(171500 cm³/x²)

A =  x² + 4 × 171500 cm³/x

A =  x² + 686000 cm³/x

To find the value of x that minimizes A, we differentiate A with respect to x

So, dA/dx = d[x² + 686000 cm³/x]/dx

dA/dx = 2x - 686000 cm³/x²

equating this to zero, we have

2x - 686000 cm³/x² = 0

2x = 686000 cm³/x²

2x³ = 686000

x³ = 686000 cm³/2

x = ∛(343000 cm³)

x = 70 cm

To determine if this in a minimum dimension for A, we differentiate dA/dx and insert x = 70 cm

So, d²A/dx² = d[2x - 686000 cm³/x²]/dx

d²A/dx² = 2 + 3 × 686000 cm³/x³

d²A/dx² = 2 + 2058000 cm³/x³

with x = 70 cm, we have

d²A/dx² = 2 + 2058000 cm³/(70 cm)³

d²A/dx² = 2 + 2058000 cm³/343000 cm³

d²A/dx² = 2 + 6

d²A/dx² = 8 > 0

So, x = 70 is a minimum value for the dimension

Since h = 171500 cm³/x²

h = 171500 cm³/(70 cm)²

h = 171500 cm³/4900 cm²

h = 35 cm

So, the dimensions of the box that minimize the amount of material used is length of square base x = 70 cm and height of box = 35 cm

hi! i think the answer is 1/2 : )

answer: c, the green line

step-by-step explanation:

there is no black line, eliminate b.

simple interest implies periodic payments, but the red line shows no payments, eliminate d.

simple interest has equal payments. if paid say every day, the graph is about a straight line (actually a stairstep).

the yellow line is not straight, eliminate a.

check: the green line is straight, answer is c.

simple interest with positive interest rate gives a straight line sloping up to the right. green line:

g(t) = $100 + ($200-$100) ÷ (20 years) × (t years)

= $100 + (4%/year) × ($100)(t years)

incidentally, the yellow line curving upward represents continuously compounded interest

y = $100 × 2^(t years / (14 assuming the yellow line hits $200 at 14 years. 2^0 is 1, so y(0) = $100; 2^(14/14) is 2, so y(14) = $200.