Subjects algebra

Variable Relations

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Variable Relations


1. Let's analyze the given equations and expressions:\n\n- $Q = 10\sqrt{KL}$\n- $K + 4L = 16$\n- $Y = 150T^{0.5} S^{0.5}$\n- $C = 10T + 8S$\n\n2. The problem seems to involve understanding the relationships among variables $K, L, T, S$, with two functions $Q$ and $Y$, and a cost $C$. Since no specific question is asked, let's express $Q$ in terms of one variable using the constraint and simplify $Y$ and $C$ if possible.\n\n3. From the constraint: $$K + 4L = 16 \implies K = 16 - 4L.$$\n\n4. Substitute $K$ into $Q$: $$Q = 10 \sqrt{K L} = 10 \sqrt{(16 - 4L)L} = 10 \sqrt{16L - 4L^2}.$$\n\n5. Simplify inside the square root:\n$$16L - 4L^2 = 4L(4 - L).$$\n\nSo, $$Q = 10 \sqrt{4L (4 - L)} = 10 \times 2 \sqrt{L (4 - L)} = 20 \sqrt{L (4 - L)}.$$\n\n6. Therefore, $$Q = 20 \sqrt{L (4 - L)}$$ with $L$ satisfying $0 \leq L \leq 4$ to keep $K \geq 0$.\n\n7. Regarding $Y$ and $C$, if $T$ and $S$ are variables, then:\n\n- $Y = 150 \sqrt{T S}$\n- $C = 10T + 8S$\n\nThese represent a production function $Y$ depending on $T$ and $S$, and a cost function $C$ depending on $T$ and $S$.\n\n8. Without further instructions (like maximizing $Q$, $Y$, or minimizing $C$), we have simplified expressions ready for analysis or optimization as needed.