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erma4kov [3.2K]
3 years ago
7

Cara has $25 to buy dry pet food and treats for the animal shelter. A pound of dog food cost $2 and treats are $1 apiece. If she

buys 9 pounds of food, what is the greatest number of treats she can buy?
Mathematics
1 answer:
nordsb [41]3 years ago
5 0
<span>If dog food costs $2 a pound and Cara buys 9 pounds, her expenditure on dog food is 9 x 2. This makes $18. If she has a total budget of $25 and spends $18 on dog food, she has $7 left over for treats (25 - 18 = 7). At $1 each, she can buy up to 7 treats.</span>
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The Jacobian for this transformation is

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dA = dx\,dy = 12 \, du\,dv

Then the integral becomes

\displaystyle \iint_{R'} 4x^2 \, dA = 768 \iint_R u^2 \, du \, dv

where R' is the unit circle,

\dfrac{x^2}{16} + \dfrac{y^2}9 = \dfrac{(4u^2)}{16} + \dfrac{(3v)^2}9 = u^2 + v^2 = 1

so that

\displaystyle 768 \iint_R u^2 \, du \, dv = 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2 \, du \, dv

Now you could evaluate the integral as-is, but it's really much easier to do if we convert to polar coordinates.

\begin{cases} u = r\cos(\theta) \\ v = r\sin(\theta) \\ u^2+v^2 = r^2\\ du\,dv = r\,dr\,d\theta\end{cases}

Then

\displaystyle 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2\,du\,dv = 768 \int_0^{2\pi} \int_0^1 (r\cos(\theta))^2 r\,dr\,d\theta \\\\ ~~~~~~~~~~~~ = 768 \left(\int_0^{2\pi} \cos^2(\theta)\,d\theta\right) \left(\int_0^1 r^3\,dr\right) = \boxed{192\pi}

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2 years ago
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