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pav-90 [236]
3 years ago
10

Find the distance between the points (-7, 5) and (4, 1).

Mathematics
2 answers:
Alla [95]3 years ago
8 0
2 because the square root is 12 and the distance is 12 units
anygoal [31]3 years ago
7 0
\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
&({{ -7}}\quad ,&{{ 5}})\quad 
%  (c,d)
&({{ 4}}\quad ,&{{ 1}})
\end{array}\qquad 
%  distance value
d = \sqrt{({{ 4}}-{{ (-7)}})^2 + ({{ 1}}-{{ 5}})^2}
\\\\\\
d=\sqrt{(4+7)^2+(1-5)^2}
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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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