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Write two-variable inequalities: word problems

A rancher wants to build a rectangular pen for her animals. She decides that the length, l l , of one side of the pen should be at most 6060 feet, the width, w w , of one side of the pen should be at least 3030 feet, and the perimeter of the pen should be at most 200200 feet. Which of the following systems of inequalities best models the situation described?\newlineChoose 11 answer:\newline(A) {l+w≤200w≥30l≤60 \left\{\begin{array}{l}l+w \leq 200 \\ w \geq 30 \\ l \leq 60\end{array}\right. \newline(B) {l+w≥200w≤30l≥60 \left\{\begin{array}{l}l+w \geq 200 \\ w \leq 30 \\ l \geq 60\end{array}\right. \newlinec) {l+w≤100w≥30l≤60 \left\{\begin{array}{l}l+w \leq 100 \\ w \geq 30 \\ l \leq 60\end{array}\right. \newline(D) {2l+2w≥200w≥30l≤60 \left\{\begin{array}{l}2 l+2 w \geq 200 \\ w \geq 30 \\ l \leq 60\end{array}\right.
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A company manufactures cellular phones and laptop computers. The company's daily production of cellular phones, cc, must be more than 500,000500,000, and its daily production of laptop computers, ll, must be more than 300,000300,000. If the maximum capacity of the company's manufacturing center is no more than 950,000950,000 total cellular phones and laptop computers, which of the following systems of inequalities best models the situation described?\newlineChoose 11 answer:\newline(A) {c−l>950,000c≥500,000l≥300,000\begin{cases} c-l > 950,000 \\ c \geq 500,000 \\ l \geq 300,000 \end{cases}\newline(B) {c−l≤950,000c>500,000l>300,000\begin{cases} c-l \leq 950,000 \\ c > 500,000 \\ l > 300,000 \end{cases}\newline(C) {c+l<950,000c≥500,000l≥300,000\begin{cases} c+l < 950,000 \\ c \geq 500,000 \\ l \geq 300,000 \end{cases}\newline(D) {c+l≤950,000c>500,000l>300,000\begin{cases} c+l \leq 950,000 \\ c > 500,000 \\ l > 300,000 \end{cases}
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Ty tried to solve the differential equation dydx=4xey \frac{d y}{d x}=\frac{4}{x e^{y}} . This is his work:\newlinedydx=4xey \frac{d y}{d x}=\frac{4}{x e^{y}} \newlineStep 11: ∫eydy=∫4xdx \quad \int e^{y} d y=\int \frac{4}{x} d x \newlineStep 22: ey=4ln⁡∣x∣+C \quad e^{y}=4 \ln |x|+C \newlineStep 33: ln⁡(ey)=ln⁡(4ln⁡∣x∣+C) \quad \ln \left(e^{y}\right)=\ln (4 \ln |x|+C) \newlineStep 44: y=ln⁡(4ln⁡∣x∣+C) \quad y=\ln (4 \ln |x|+C) \newlineIs Ty's work correct? If not, what is his mistake?\newlineChoose 11 answer:\newline(A) Ty's work is correct.\newline(B) Step 11 is incorrect. The separation of variables was done incorrectly.\newline(C) Step 33 is incorrect. The right-hand side of the equation should be ln⁡(4ln⁡∣x∣)+C \ln (4 \ln |x|)+C .\newline(D) Step 44 is incorrect. The left-hand side of the equation should be ∣y∣ |y| .
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Eleanor tried to find all the equations of vertical lines tangent to the curve given by\newline3x2+24x−y2+2y=−44.  3 x^{2}+24 x-y^{2}+2 y=-44 \text {. } \newlineThis is her solution:\newlineStep 11: Finding an expression for dydx \frac{d y}{d x} .\newlinedydx=y−13(x+4) \frac{d y}{d x}=\frac{y-1}{3(x+4)} \newlineStep 22: Forming a system of equations.\newline{3x2+24x−y2+2y=−443(x+4)=0y−1≠0 \left\{\begin{array}{l} 3 x^{2}+24 x-y^{2}+2 y=-44 \\ 3(x+4)=0 \\ y-1 \neq 0 \end{array}\right. \newlineStep 33: Solving the system.\newlineThere are no solutions to the system of equations, so there are no vertical lines tangent to the graph.\newlineIs Eleanor's solution correct? If not, at which step did she make a mistake?\newlineChoose 11 answer:\newline(A) The solution is correct.\newline(B) Step 11 is incorrect.\newline(C) Step 2 \mathbf{2} is incorrect.\newline(D) Step 33 is incorrect.
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