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3 years ago

5

On two tests so far this year, a student received a 78 and a 93. The student wants an average between 80 and 90 inclusive. What

are all of the possible scores for the third test so that the student falls within this average?

1 answer:

Mariulka [41]3 years ago

8 0

<span>240 <= 171 + x <= 270
and now subtract 171 from all and get
69 <= x <= 99
So any score from 69 to 99 will do.</span>

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A box with a square base and open top must have a volume of 296352 c m 3 . We wish to find the dimensions of the box that minimi

mestny [16]

Answer:

Base Length of 84cm
Height of 42 cm.

Step-by-step explanation:

Given a box with a square base and an open top which must have a volume of 296352 cubic centimetre. We want to minimize the amount of material used.

Step 1:

Let the side length of the base =x

Let the height of the box =h

Since the box has a square base

Volume, $V=x^2h=296352$

$h=\dfrac{296352}{x^2}$

Surface Area of the box = Base Area + Area of 4 sides

$A(x,h)=x^2+4xh\\$Substitute h=\dfrac{296352}{x^2}\\A(x)=x^2+4x\left(\dfrac{296352}{x^2}\right)\\A(x)=\dfrac{x^3+1185408}{x}$

Step 2: Find the derivative of A(x)

$If\:A(x)=\dfrac{x^3+1185408}{x}\\A'(x)=\dfrac{2x^3-1185408}{x^2}$

Step 3: Set A'(x)=0 and solve for x

$A'(x)=\dfrac{2x^3-1185408}{x^2}=0\\2x^3-1185408=0\\2x^3=1185408\\$Divide both sides by 2\\x^3=592704\\$Take the cube root of both sides\\x=\sqrt[3]{592704}\\x=84$

Step 4: Verify that x=84 is a minimum value

We use the second derivative test

$A''(x)=\dfrac{2x^3+2370816}{x^3}\\$When x=84$\\A''(x)=6$

Since the second derivative is positive at x=84, then it is a minimum point.

Recall:

$h=\dfrac{296352}{x^2}=\dfrac{296352}{84^2}=42$

Therefore, the dimensions that minimizes the box surface area are:

Base Length of 84cm
Height of 42 cm.

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