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Elodia [21]
2 years ago
10

find the length and width of a rectangle that has an area of 968 square feet and whose perimeter is a minimum

Mathematics
1 answer:
andrezito [222]2 years ago
3 0
The question isn't finished...
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Find the measure of angle b (image below)
prohojiy [21]

Answer:

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3 is less than the sum of a number s and 4
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3 is less than the sum of any positive number added with 4
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Zadanie optymalizacyjne z matematyki. Proszę o rozwiązanie i wyjaśnienie
Yakvenalex [24]

Volume of the pyramid:

V=\dfrac{s^2h}3

Perimeter of the cross-section:

40=\sqrt2\,s+2\sqrt{\dfrac{s^2}2+h^2}=\sqrt2\left(s+\sqrt{s^2+2h^2}\right)

\implies h=\sqrt{\dfrac{(20\sqrt2-s)^2-s^2}2}=\sqrt{400-20\sqrt2\,s}

Area of the cross-section:

P=\dfrac12(\sqrt2\,s)h=\dfrac{sh}{\sqrt2}

\implies P=\dfrac{s\sqrt{400-20\sqrt2\,s}}{\sqrt2}=s\sqrt{200-10\sqrt2\,s}

First derivative test:

\dfrac{\mathrm dP}{\mathrm ds}=\dfrac{20\sqrt2-3s}{\sqrt{4-\dfrac{\sqrt2}5s}}=0\implies s=\dfrac{20\sqrt2}3

Then the height of the cross-section/pyramid is

h=\sqrt{400-20\sqrt2\,s}=\dfrac{20}{\sqrt3}

The volume of the pyramid that maximizes the cross-sectional area P is

V=\dfrac{\left(\frac{20\sqrt2}3\right)^2\frac{20}{\sqrt3}}3=\dfrac{16000}{27\sqrt3}

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3 years ago
Question about dilation attached in images.
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This is an enlarged dilation
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108+621 is the same as 100+
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Answer:

108+621= 100+629 hhh grade 2?

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