I will assume you mistyped this question. For y = -1/16x^2 + 4x + 3, the answers to this question are
a) 3 feet
b) 67 feet
c) 64.741 feet

For a) we note that at x = 0, that is the instant where the ball leaves the hand. y(0) = 0.

For b), we find the vertex of y = -1/16x^2 + 4x + 3
                               y = -1/16x^2 + 4x + 3
                               y = -1/16(x^2 - 64x) + 3
                               y = -1/16(x^2 - 64x + 1024 - 1024) + 3
                               y = -1/16((x-32)^2 - 1024) + 3
                               y = -1/16(x-32)^2 + 64 + 3
                               y = -1/16(x-32)^2 + 67
The vertex is at (32,67) so 67 is the maximum height.

For c), we find the x-intercepts with the quadratic formula on
y = -1/16x^2 + 4x + 3=0:
                     x = [ -b ± √b^2 - 4ac ] / (2a)<span>
                     x = [ -4 ± √4^2 - 4(-1/16)(3) ] / (2(-1/16))
   x = -0.741, 64.741
Only the positive solution, so 64.741 feet         </span>

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3 years ago

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A rectangular box is designed to have a square base and an open top. The volume is to be 500in.3 What is the minimum surface are

mr_godi [17]

The minimum surface area that such a box can have is 380 square

<h3>How to determine the minimum surface area such a box can have?</h3>

Represent the base length with x and the bwith h.

So, the volume is

V = x^2h

This gives

x^2h = 500

Make h the subject

h = 500/x^2

The surface area is

S = 2(x^2 + 2xh)

Expand

S = 2x^2 + 4xh

Substitute h = 500/x^2

S = 2x^2 + 4x * 500/x^2

Evaluate

S = 2x^2 + 2000/x

Differentiate

S' = 4x - 2000/x^2

Set the equation to 0

4x - 2000/x^2 = 0

Multiply through by x^2

4x^3 - 2000 = 0

This gives

4x^3= 2000

Divide by 4

x^3 = 500

Take the cube root

x = 7.94

Substitute x = 7.94 in S = 2x^2 + 2000/x

S = 2 * 7.94^2 + 2000/7.94

Evaluate

S = 380

Hence, the minimum surface area that such a box can have is 380 square